◆ QA · Algebra

Algebra, formulas + CAT PYQs

Equations & polynomials, quadratics (roots, discriminant, nature), inequalities & modulus, functions & graphs, logarithms & indices, surds, and progressions (AP/GP/HP). The single largest QA area after arithmetic, ~22 questions per CAT.

30formulas

209CAT PYQs

★★★priority

Formula Sheet CAT PYQs

Linear & Simultaneous Quadratics Inequalities & Modulus Functions & Graphs Logarithms Surds & Indices Progressions & Series Maxima-Minima

Formula & Concept Sheet

A-to-Z. Everything you need for this chapter, distilled from the Revision Notes.

1Polynomials & zeroes

Plain English: a polynomial is just a sum of x-powers; its "zeroes" are the x-values that make it equal 0.
A polynomial of degree n: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
k is a zero of p(x) if p(k) = 0. Zeroes are the x-coordinates where y = p(x) cuts the x-axis.
Max zeroes = degree: linear → 1, quadratic → 2, cubic → 3, degree n → n.
e.g. p(x) = x² − 9 has degree 2, so at most 2 zeroes: x = 3 and x = −3.

2Sum & product of roots

Plain English: you can read the sum and product of the roots straight off the coefficients, no need to solve.
Quadratic ax²+bx+c: α + β = −b/a, αβ = c/a
Cubic ax³+bx²+cx+d: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a
Build the equation: x² − (sum)x + (product) = 0.
e.g. x² − 5x + 6 = 0: sum = 5, product = 6 ⇒ roots 2 and 3 (2+3=5, 2×3=6).

3Discriminant & nature of roots

Plain English: the discriminant D is a single number that tells you how many real roots a quadratic has, before solving.
For ax²+bx+c (a≠0): D = b² − 4ac
D > 0 → two distinct real roots; D = 0 → equal real roots; D < 0 → no real roots (complex).
D a perfect square (a,b,c rational) → roots are rational.
Roots: x = (−b ± √D)/2a
e.g. x² + x + 1: D = 1 − 4 = −3 < 0 ⇒ no real roots.

4Sum of squares of roots (trick)

Plain English: you can get α²+β² from the sum and product alone, never solve for the roots first.
α² + β² = (α+β)² − 2αβ
To minimise a sum-of-squares-of-roots expression in a parameter, complete the square, minimum is at the vertex.
e.g. if α+β = 3 and αβ = 2, then α²+β² = 9 − 4 = 5.

5Algebraic identities

Plain English: memorised expand/factor templates that turn ugly expressions into products (or vice-versa) instantly.
(a±b)² = a² ± 2ab + b²
a² − b² = (a+b)(a−b)
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
a³ ± b³ = (a±b)(a² ∓ ab + b²)
a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca)
e.g. 97×103 = (100−3)(100+3) = 100² − 3² = 9991.

6Linear equations in two variables

Plain English: comparing the coefficient ratios tells you whether two lines cross once, never, or lie on top of each other.
a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.
Unique solution (intersecting): a₁/a₂ ≠ b₁/b₂
No solution (parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Infinite solutions (coincident): a₁/a₂ = b₁/b₂ = c₁/c₂
e.g. x+y=2 and 2x+2y=5: ratios 1/2 = 1/2 ≠ 2/5 ⇒ parallel, no solution.

7Inequalities, basic rules

Plain English: inequalities behave like equations, except multiplying or dividing by a negative reverses the arrow.
Adding/subtracting keeps direction; multiplying by a negative flips the sign.
If X > Y > 0 then 1/X < 1/Y.
For x > 0: x + 1/x ≥ 2 (equality at x = 1).
e.g. −2x > 6 ⇒ divide by −2 and flip ⇒ x < −3.

8Quadratic inequalities

Plain English: factor it, then "< 0" means between the roots and "> 0" means outside the roots.
(x−m)(x−n) < 0, n > m ⇒ m < x < n (between the roots).
(x−m)(x−n) > 0 ⇒ x < m or x > n (outside the roots).
Sign-of-product / wavy-curve method handles higher degree.
e.g. x² − 5x + 6 < 0 ⇒ (x−2)(x−3) < 0 ⇒ 2 < x < 3.

9Modulus (absolute value)

Plain English: |x| is the distance of x from 0, so it strips the sign and is never negative.
|x| = max(x, −x); −|x| ≤ x ≤ |x|.
|a+b| ≤ |a|+|b| and |a|−|b| ≤ |a−b|; |ab| = |a||b|.
|x| ≤ k ⇒ −k ≤ x ≤ k. |x| ≥ k ⇒ x ≥ k or x ≤ −k.
|f| + |g| = |f+g| only when f, g have the same sign.
e.g. |x| ≤ 3 ⇒ −3 ≤ x ≤ 3; |x − 4| = 2 ⇒ x = 6 or x = 2.

10AM-GM-HM inequality

Plain English: for positive numbers the plain average is always ≥ the geometric average, the go-to tool for "find the minimum".
For positive reals: AM ≥ GM ≥ HM, equality when all equal.
Two numbers: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).
AM × HM = GM²
If a₁a₂…aₙ = 1 then a₁+a₂+…+aₙ ≥ n.
e.g. for a = 2, b = 8: AM = 5 ≥ GM = √16 = 4. ✓

11Maxima & minima of a quadratic

Plain English: a parabola's turning point is at x = −b/2a; that's where the min (opens up) or max (opens down) lives.
ax²+bx+c: vertex at x = −b/2a; extreme value = (4ac − b²)/4a = −D/4a.
a > 0 → opens up → minimum; a < 0 → opens down → maximum.
min/max of max-of-two / min-of-two lines occurs where the two graphs intersect.
e.g. x² − 6x + 5: vertex at x = 3, minimum value = 9 − 18 + 5 = −4.

12Functions, domain, range, even/odd

Plain English: domain is what you may feed in, range is what comes out; even/odd describe the graph's symmetry.
Domain = allowed inputs; range = resulting outputs.
Even: f(−x) = f(x) (graph symmetric about y-axis), e.g. x², |x|.
Odd: f(−x) = −f(x) (symmetric about origin), e.g. x³, 1/x.
Inverse exists only if f is one-to-one.
e.g. f(x) = x³ is odd: f(−2) = −8 = −f(2). ✓

13Functional equations

Plain English: the form of a functional rule reveals the function, "turns + into ×" means exponential, etc.
f(x+y) = f(x)·f(y) ⇒ exponential type, f(x) = aˣ.
f(xy) = f(x)·f(y) ⇒ power/multiplicative; f(1) = 1.
If f(a+x) = f(a−x), the graph is symmetric about x = a; roots pair around a (sum of 4 roots = 4a).
e.g. f(x+y) = f(x)f(y) with f(1) = 3 ⇒ f(2) = f(1)² = 9.

14Graph shifting

Plain English: changes outside f() move the graph vertically; changes inside f() move it horizontally (and oppositely).
f(x)+c → shift up c; f(x)−c → shift down c.
f(x+c) → shift left c; f(x−c) → shift right c.
−f(x) → reflect in x-axis; f(−x) → reflect in y-axis.
e.g. y = (x−2)² is y = x² shifted 2 units right.

15Logarithm, definition

Plain English: log_b x just asks "what power of b gives x?", it's the inverse of raising to a power.
y = log_b x ⇔ x = bʸ (b > 0, b ≠ 1, x > 0).
log_a a = 1; log_a 1 = 0; a^(log_a m) = m.
e.g. log₂8 = 3 because 2³ = 8.

16Logarithm laws

Plain English: logs turn multiplication into addition, division into subtraction, and powers into multipliers.
log_a(xy) = log_a x + log_a y
log_a(x/y) = log_a x − log_a y
log_a(xᵐ) = m·log_a x
log_(aⁿ)(xᵐ) = (m/n)·log_a x
Change of base: log_a x = (log x)/(log a); log_a x = 1/log_x a
e.g. log₂40 = log₂(8×5) = log₂8 + log₂5 = 3 + log₂5.

17Indices (laws of exponents)

Plain English: same base, add exponents when multiplying, subtract when dividing, multiply when raising a power to a power.
pᵐ·pⁿ = pᵐ⁺ⁿ; pᵐ/pⁿ = pᵐ⁻ⁿ; (pᵐ)ⁿ = pᵐⁿ
pⁿ·qⁿ = (pq)ⁿ; (p/q)ⁿ = pⁿ/qⁿ
p⁻ⁿ = 1/pⁿ; p⁰ = 1; p^(1/n) = ⁿ√p
e.g. 2³·2⁴ = 2⁷ = 128; 8^(2/3) = (∛8)² = 2² = 4.

18Surds & rationalisation

Plain English: a surd is an unresolved root like √2; "rationalising" clears it from a denominator using the conjugate.
√(ab) = √a·√b; √(a/b) = √a/√b.
Rationalise a/(b+√c) by multiplying top & bottom by the conjugate (b−√c).
If a+√b is a root of a rational quadratic, so is its conjugate a−√b.
e.g. 1/(√3 − 1) × (√3 + 1)/(√3 + 1) = (√3 + 1)/2.

19Arithmetic Progression (AP)

Plain English: an AP adds the same step d each time; its sum is just "how many terms × the average of first and last".
Constant difference d. nth term: Tₙ = a + (n−1)d
Sum: Sₙ = n/2 · [2a + (n−1)d] = n/2 · (first + last)
Arithmetic mean of a, b: A = (a+b)/2. Middle term = average of an odd count of AP terms.
e.g. 2, 5, 8, …: T₄ = 2 + 3×3 = 11; sum of first 4 = 4/2·(2+11) = 26.

20Geometric Progression (GP)

Plain English: a GP multiplies by the same ratio r each time; if |r| < 1 the infinite sum settles to a finite value.
Constant ratio r. nth term: Tₙ = a·rⁿ⁻¹
Sum: Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1.
Infinite sum (|r| < 1): S∞ = a/(1 − r)
Geometric mean: G = √(ab).
e.g. 1 + ½ + ¼ + … = 1/(1 − ½) = 2.

21Harmonic Progression (HP)

Plain English: an HP is just an AP flipped, take reciprocals and you're back to a normal AP.
a, b, c… in HP ⇔ 1/a, 1/b, 1/c… in AP.
Harmonic mean of a, b: H = 2ab/(a+b)
nth term of HP = 1/(nth term of the corresponding AP).
e.g. 1, ½, ⅓, ¼ is an HP (reciprocals 1, 2, 3, 4 form an AP).

22Standard summation formulas

Plain English: ready-made closed forms for adding up the first n numbers, their squares, and their cubes.
Σn = n(n+1)/2
Σn² = n(n+1)(2n+1)/6
Σn³ = [n(n+1)/2]²
Telescoping: 1/(k·(k+1)) = 1/k − 1/(k+1).
e.g. 1 + 2 + … + 10 = 10×11/2 = 55.

23Common terms of two APs

Plain English: numbers shared by two APs themselves form an AP whose step is the LCM of the two steps.
Common terms of two APs form a new AP with common difference = LCM of the two differences.
Find the first common term, then count multiples of the LCM up to the smaller upper limit.
e.g. 2,5,8,… and 3,7,11,…: first common term 11, new step = LCM(3,4) = 12 ⇒ 11, 23, 35, …

24Recurrence & tₙ from Sₙ

Plain English: if you know the running total Sₙ, each term is just this total minus the previous total.
If Sₙ given: aₙ = Sₙ − Sₙ₋₁ (and a₁ = S₁).
Alternating-sum sequences: subtract consecutive defining equations to isolate a term.
e.g. Sₙ = n² ⇒ a₅ = S₅ − S₄ = 25 − 16 = 9.

25Integer / Diophantine solutions

Plain English: once you spot one whole-number solution, all the rest come by stepping x and y in fixed jumps.
ax + by = c with one integer solution (x₀, y₀): all others are x₀ + (b/g)t, y₀ − (a/g)t, where g = gcd(a,b).
Bound the count using the given ranges on x and y.
e.g. 2x + 3y = 12: (x,y) = (3,2) works; next is (0,4), then (6,0), x jumps by 3, y by 2.

26When does Aᴮ = 1?

Plain English: a power equals 1 in exactly three situations, check all three or you'll miss cases.
Base = 1 (any exponent), or
Exponent = 0 (base ≠ 0), or
Base = −1 with an even exponent.
e.g. (−1)⁴ = 1 (base −1, even power); 7⁰ = 1 (zero power); 1⁹⁹ = 1 (base 1).

27Three terms in AP / GP

Plain English: centering three terms on a middle value makes their sum (AP) or product (GP) collapse to one symbol.
Three in AP: take a−d, a, a+d (their sum = 3a).
Three in GP: take a/r, a, ar (product = a³).
Three consecutive integers as roots: n−1, n, n+1.
e.g. three numbers in AP summing to 18 ⇒ middle = 6, so 6−d, 6, 6+d.

28|x − a| as distance (modulus sums)

Plain English: read |x−a| as "distance from a", and sums of such distances are smallest when x sits among the points.
|x−a| = distance of x from a on the number line.
|x−p|+|x−q| is minimised for any x between p and q; minimum value = |p−q|.
|x−p| = |x−q| at the midpoint x = (p+q)/2.
e.g. |x−2| + |x−7| ≥ 5, achieved for any x in [2, 7].

29Sum of squares identity trick

Plain English: squares can't be negative, so if a bunch of squares add to 0 every single one must be 0.
If a sum of squares equals 0, each square = 0: e.g. (x−2y)² + (y−z)² = 0 ⇒ x = 2y and y = z.
Group given expressions into perfect squares to pin exact values.
e.g. (a−3)² + (b+1)² = 0 forces a = 3 and b = −1.

30Cauchy / vector identity

Plain English: this identity links two "sum-of-squares" products to two cross-terms, handy when three of the four pieces are given.
(a²+b²)(x²+y²) = (ax+by)² + (ay−bx)².
Useful when given a²+b², x²+y² and ax+by to find ay−bx.
e.g. (1²+2²)(3²+4²) = 5·25 = 125 = 11² + 2² = (1·3+2·4)² + (1·4−2·3)².

Linear & Simultaneous Equations · 29 CAT PYQs

Linear & Simultaneous Equations

ModerateCAT 1999

The number of positive integer valued pairs (x, y) satisfying 4x − 17y = 1 and x ≤ 1000 is

(1) 59
(2) 57
(3) 55
(4) 58

Show solution

(1) 59. A base solution is (x, y) = (13, 3); general x increases by 17 each step. Counting x ≤ 1000 from x = 13 onward in steps of 17 gives 59 pairs.

ModerateCAT 2000

If x³ − ax² + bx − a = 0 has three real roots, then it must be the case that

(1) b = 1
(2) b ≠ 1
(3) a = 1
(4) a ≠ 1

Show solution

(2) b ≠ 1. Test b = 1: the factorisation becomes (x − a)(x² + 1), and x² + 1 = 0 has no real roots, so only one real root, contradiction. Hence b ≠ 1.

ModerateCAT 2002

If x² + 5y² + z² = 2y(2x + z), then which of the following statements are necessarily true? I. x = 2y II. x = 2z III. 2x = z

(1) Only I
(2) Only II
(3) Only III
(4) Only I and II

Show solution

(4) Only I and II. Rearrange: (x² − 4xy + 4y²) + (y² − 2yz + z²) = 0 ⇒ (x − 2y)² + (y − z)² = 0. Each square is 0 ⇒ x = 2y and y = z ⇒ x = 2z. So I and II hold.

HardCAT 2003

Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ≠ 0? x + 2y − 3z = p, 2x + 6y − 11z = q, x − 2y + 7z = r

(1) 5p − 2q − r = 0
(2) 5p + 2q + r = 0
(3) 5p + 2q − r = 0
(4) 5p − 2q + r = 0

Show solution

(1) 5p − 2q − r = 0. Eliminating z from pairs of equations gives 5x+4y = 11p−3q and 5x+4y = (7q+11r)/5. For consistency these must be equal: 11p−3q = (7q+11r)/5 ⇒ 5p − 2q − r = 0.

ModerateCAT 2003

The number of roots common between the two equations x³ + 3x² + 4x + 5 = 0 and x³ + 2x² + 7x + 3 = 0 is

(1) 0
(2) 1
(3) 2
(4) 3

Show solution

(1) 0. Subtracting the equations: x² − 3x + 2 = 0 ⇒ x = 1 or 2. Neither value satisfies either original cubic, so there are no common roots.

ModerateCAT 2003

A test has 50 questions. A student scores 1 mark for a correct answer, −1/3 for a wrong answer, and −1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than

(1) 6
(2) 12
(3) 3
(4) 9

Show solution

(3) 3. With R right, W wrong, N not attempted: R + W + N = 50 and R − W/3 − N/6 = 32. Solving gives R = (242 + W)/7, so 242 + W must be divisible by 7; the smallest such W is 3.

HardCAT 2005

For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive? x² − y² = 0, (x − k)² + y² = 1

(1) 2
(2) 0
(3) √2
(4) −√2

Show solution

(3) √2. x² = y²; substitute: 2x² − 2kx + (k²−1) = 0. Unique solution ⇒ discriminant 0: 4k² − 8(k²−1) = 0 ⇒ k² = 2 ⇒ k = ±√2. For a positive x, k = √2.

ModerateCAT 2006

The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x ≤ y is:

(1) 7
(2) 13
(3) 14
(4) 18

Show solution

(2) 13. y = 40 − 2x > 0 and x ≤ y. Valid x runs from 1 to 13 (at x = 13, y = 14 ≥ x; at x = 14, y = 12 < x). So 13 solutions.

HardCAT 2019

The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is:

(1) 58
(2) 85
(3) 50
(4) 95

Show solution

(3) 50. xy = 616. (x³−y³)/(x−y)³ = 157/3 ⇒ 3(x²+xy+y²) = 157(x²+y²−2xy) ⇒ 154(x+y)² = 625xy = 625·616 ⇒ (x+y)² = 625·4 = 2500 ⇒ x+y = 50.

ModerateCAT 2019TITA

The number of solutions to the equation |x|(6x² + 1) = 5x² is

Show solution

5. x = 0 works. Otherwise divide by |x|: 6x² + 1 = 5|x|. Let |x| = y: 6y² − 5y + 1 = 0 ⇒ y = 1/2 or 1/3, giving x = ±1/2, ±1/3. Total with 0 = 5 solutions.

HardCAT 2019

Let a, b, x, y be real numbers such that a² + b² = 81, x² + y² = 121, and ax + by = 99. If k = ay − bx, then

(1) 0 < k ≤ 5/13
(2) k > 5/13
(3) k = 5/13
(4) k = 0

Show solution

(4) k = 0. By the identity (a²+b²)(x²+y²) = (ax+by)² + (ay−bx)²: 81·121 = 99² + k². But 81·121 = 9801 = 99². So k² = 0 ⇒ k = 0.

ModerateCAT 2017

The number of solutions (x, y, z) to the equation x − y − z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

(1) 101
(2) 99
(3) 87
(4) 105

Show solution

(2) 99. x = 25 + y + z with 2 ≤ y+z ≤ 15 (since x ≤ 40) and y, z ≥ 1. Counting valid (y, z) pairs: y = 1,2,3 each give 12 values of z; y = 4 gives 11; …down to y = 12 gives 3. Total = 99.

EasyCAT 2020

Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if

(1) |k| ≠ 2
(2) k = 2
(3) k ≠ 2
(4) |k| = 2

Show solution

(1) |k| ≠ 2. Unique solution requires k/4 ≠ 1/k ⇒ k² ≠ 4 ⇒ k ≠ ±2 ⇒ |k| ≠ 2.

ModerateCAT 2020TITA

The number of pairs of integers (x, y) satisfying x ≥ y ≥ −20 and 2x + 5y = 99 is:

Show solution

17. 2x even ⇒ 5y odd ⇒ y odd. Minimum y = −19; x ≥ y caps y at 13. Odd y from −19 to 13 gives 17 values.

ModerateCAT 2020TITA

If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

Show solution

23. Adding the equalities: x + y = 2z − 10 < z + 5 ⇒ z < 15 ⇒ z_max = 14, giving x = 5, y = 13. Then 2x + y = 10 + 13 = 23.

ModerateCAT 2021 · Slot 2

Consider the pair of equations: x² − xy − x = 22 and y² − xy + y = 34. If x > y, then x − y equals

(1) 7
(2) 4
(3) 6
(4) 8

Show solution

(4) 8. Add the two equations: x² − 2xy + y² − x + y = 56 ⇒ (x−y)² − (x−y) = 56. Let t = x−y: t² − t − 56 = 0 ⇒ (t−8)(t+7) = 0 ⇒ t = 8 (since x > y).

ModerateCAT 2022 · Slot 1TITA

For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is:

Show solution

34. From xy + yz = y(x+z) = 19 (prime) ⇒ y = 1, x + z = 19. Then yz + xz = z(x+1)... using z(x+y) = 51 with y = 1: z(x+1) = 51. With x + z = 19, solving gives x = 2, z = 17 ⇒ xyz = 2·1·17 = 34.

HardCAT 2022 · Slot 2

Let r and c be real numbers. If r and −r are roots of 5x³ + cx² − 10x + 9 = 0, then c equals:

(1) 4
(2) −4
(3) −9/2
(4) 9/2

Show solution

(3) −9/2. Let third root α. Sum of roots = α = −c/5. Product = α(−r²) = −9/5. Sum of products pairwise = −r² = −10/5 = −2 ⇒ r² = 2. From product: α·(−2) = −9/5... solving the relations gives c = −9/2.

HardCAT 2023 · Slot 1TITA

The equation x³ + (2r + 1)x² + (4r − 1)x + 2 = 0 has −2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

Show solution

2. Dividing out (x+2) leaves a quadratic in x; requiring its discriminant ≥ 0 gives a bound on r, whose smallest non-negative integer value is 2.

HardCAT 2023 · Slot 2

If p² + q² − 29 = 2pq − 20 = 52 − 2pq, then the difference between the maximum and minimum possible value of (p³ − q³) is

(1) 243
(2) 378
(3) 189
(4) 486

Show solution

(2) 378. From 2pq − 20 = 52 − 2pq ⇒ pq = 18; and p²+q² − 29 = 2pq − 20 ⇒ (p−q)² = 9 ⇒ p−q = ±3. With pq = 18: (p, q) = (6, 3) or (−6, −3). p³−q³ = 189 or −189. Difference = 378.

ModerateCAT 2023 · Slot 3

For some real numbers a and b, the system of equations x + y = 4 and (a + 5)x = (b² − 15)y = 8b has infinitely many solutions for x and y. Then, the maximum possible value of ab is

(1) 33
(2) 55
(3) 15
(4) 25

Show solution

(1) 33. Matching coefficients for infinitely many solutions ties a and b through 8b and 4(a+5) = 4(b²−15) = 8b. Solving and maximising ab gives 33.

CAT 2024 & 2025, recent

HardCAT 2024 · Slot 1

The sum of all real values of k for which (1/8)^k × (1/32768)^(1/3) = (1/8) × (1/32768)^(1/k), is

(A) 2/3
(B) 4/3
(C) −2/3
(D) −4/3

Show solution

ModerateCAT 2024 · Slot 1 TITA

If x is a positive real number such that 4 log₁₀x + 4 log₁₀₀x + 8 log₁₀₀₀x = 13, then the greatest integer not exceeding x, is

Show solution

31. Converting bases: (4 + 2 + 8/3)·log₁₀x = (26/3)·log₁₀x = 13 ⇒ log₁₀x = 1.5 ⇒ x = 10^1.5 ≈ 31.62, so ⌊x⌋ = 31.

ModerateCAT 2024 · Slot 3 TITA

The number of distinct integer solutions (x, y) of the equation |x + y| + |x − y| = 2, is

Show solution

8. The equation means max(|x|, |y|) = 1, i.e. the integer points on the boundary of the square [−1,1]², there are 8 such lattice points.

ModerateCAT 2025 · Slot 1

If a − 6b + 6c = 4 and 6a + 3b − 3c = 50, where a, b and c are real numbers, the value of 2a + 3b − 3c is

(A) 18
(B) 20
(C) 15
(D) 14

Show solution

(A) 18. Let u = 3b − 3c. Eq2: 6a + u = 50; eq1: a − 2u = 4. Solving, a = 8, u = 2. So 2a + 3b − 3c = 2a + u = 16 + 2 = 18.

HardCAT 2025 · Slot 1

The number of distinct integers n for which log₍₁/₄₎(n² − 7n + 11) > 0, is

(A) infinite
(B) 0
(C) 2
(D) 1

Show solution

(B) 0. Base 1/4 < 1, so log > 0 needs 0 < n²−7n+11 < 1. The quadratic equals an integer for integer n, so it cannot lie strictly between 0 and 1. Hence no integer works.

ModerateCAT 2025 · Slot 1

In the set of consecutive odd numbers {1, 3, 5, …, 57}, there is a number k such that the sum of all the elements less than k is equal to the sum of all the elements greater than k. Then, k equals

(A) 41
(B) 39
(C) 37
(D) 43

Show solution

(A) 41. The 29 odd terms sum to 29² = 841. For k = 41, terms below sum to 400 and above sum to 400. ✓

HardCAT 2025 · Slot 1

For any natural number k, let aₖ = 3ᵏ. The smallest natural number m for which (a₁)¹ × (a₂)² × … × (a₂₀)²⁰ < a₂₁ × a₂₂ × … × a₍₂₀₊ₘ₎, is

(A) 59
(B) 56
(C) 58
(D) 57

Show solution

(C) 58. LHS exponent = Σk·k for k=1..20 = Σk² = 2870. RHS exponent = Σ(20+j) for j=1..m = 20m + m(m+1)/2. Need 20m + m(m+1)/2 > 2870 → smallest m = 58.

ModerateCAT 2025 · Slot 3 TITA

In an arithmetic progression, if the sum of the fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of the first five terms is

Show solution

65. 4th+7th+10th = 3(a+6d) = 99 → a + 6d = 33. S₁₄ = 7(2a+13d) = 497 → 2a + 13d = 71. Solving: d = 5, a = 3. S₅ = 5/2(6 + 20) = 65.

Quadratic Equations · 29 CAT PYQs

Quadratic Equations

ModerateCAT 1996

Given the quadratic equation x² − (A − 3)x − (A − 2), for what value of A will the sum of the squares of the roots be zero?

(1) − 2
(2) 3
(3) 6
(4) None of these

Show solution

(4) None of these. α+β = A−3, αβ = −(A−2). α²+β² = (A−3)² + 2(A−2) = A² − 4A + 5 = 0 has no real solution for A. So none of the choices works.

ModerateCAT 1997

If the roots x₁ and x₂ of the quadratic equation x² − 2x + c = 0 also satisfy the equation 7x₂ − 4x₁ = 47, then which of the following is true?

(1) c = − 15
(2) x₁ = − 5, x₂ = 3
(3) x₁ = 4.5, x₂ = − 2.5
(4) None of these

Show solution

(1) c = −15. Sum of roots x₁+x₂ = 2. Solving with 7x₂ − 4x₁ = 47 gives x₁ = −3, x₂ = 5. Product of roots = (−3)(5) = −15 = c.

ModerateCAT 2003

Let p and q be the roots of the quadratic equation x² − (α − 2)x − α − 1 = 0. What is the minimum possible value of p² + q²?

(1) 0
(2) 3
(3) 4
(4) 5

Show solution

(4) 5. p+q = α−2, pq = −α−1. p²+q² = (p+q)² − 2pq = (α−2)² + 2(α+1) = α² − 2α + 5 = (α−1)² + 5. Minimum is 5 (at α = 1).

ModerateCAT 2008

If the roots of the equation x³ − ax² + bx − c = 0 are three consecutive integers, then what is the smallest possible value of b?

(1) − 1/√3
(2) − 1
(3) 0
(4) 1

Show solution

(2) −1. Roots n−1, n, n+1. b = sum of products in pairs = (n−1)n + n(n+1) + (n−1)(n+1) = 3n² − 1. Minimum at n = 0 gives b = −1.

ModerateCAT 2017

The minimum possible value of the sum of the squares of the roots of the equation x² + (a + 3)x − (a + 5) = 0 is

(1) 1
(2) 2
(3) 3
(4) 4

Show solution

(3) 3. Sum = −(a+3), product = −(a+5). Sum of squares = (a+3)² + 2(a+5) = a² + 8a + 19 = (a+4)² + 3. Minimum is 3.

ModerateCAT 2018TITA

If a and b are integers such that 2x² − ax + 2 > 0 and x² − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is

Show solution

36. 2x²−ax+2 > 0 ⇒ a² < 16 ⇒ a ≤ 3. x²−bx+8 ≥ 0 ⇒ b² ≤ 32 ⇒ −5 ≤ b ≤ 5. Max of 2a − 6b = 2(3) − 6(−5) = 36.

HardCAT 2019

The product of the distinct roots of |x² − x − 6| = x + 2 is

(1) − 16
(2) − 4
(3) − 24
(4) − 8

Show solution

(1) −16. Case x²−x−6 = x+2 ⇒ x²−2x−8 = 0 ⇒ x = 4, −2. Case x²−x−6 = −(x+2) ⇒ x² = 4 ⇒ x = ±2. Distinct roots: {4, −2, 2}. Product = 4 × (−2) × 2 = −16.

ModerateCAT 2019

The quadratic equation x² + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b² + c?

(1) 3721
(2) 361
(3) 427
(4) 549

Show solution

(4) 549. Sum = 7a = −b ⇒ b² = 49a². Product = 12a² = c. b² + c = 61a². Among the options only 549 = 61 × 9 = 61 × 3². So b² + c = 549.

ModerateCAT 2020TITA

The number of distinct real roots of the equation (x + 1/x)² − 3(x + 1/x) + 2 = 0 equals:

Show solution

1. Let y = x + 1/x: y² − 3y + 2 = 0 ⇒ y = 1 or 2. But |x + 1/x| ≥ 2 for real x, so y = 1 is impossible; only y = 2 ⇒ x = 1. One distinct real root.

ModerateCAT 2020

Let f(x) = x² + ax + b and g(x) = f(x + 1) − f(x − 1). If f(x) ≥ 0 for all real x, and g(20) = 72, then the smallest possible value of b is:

(1) 1
(2) 4
(3) 0
(4) 16

Show solution

(2) 4. g(20) = f(21) − f(19) = 72 ⇒ 4a + 80 = 72... solving gives a = −4. f(x) ≥ 0 needs D ≤ 0: a² − 4b ≤ 0 ⇒ 16 ≤ 4b ⇒ b ≥ 4. Smallest b = 4.

HardCAT 2020

Let m and n be positive integers, If x² + mx + 2n = 0 and x² + 2nx + m = 0 have real roots, then the smallest possible value of m + n is:

(1) 7
(2) 6
(3) 8
(4) 5

Show solution

(2) 6. Real roots need m² ≥ 8n and (2n)² ≥ 4m ⇒ n² ≥ m. n = 1 ⇒ m ≤ 1 but m² ≥ 8 fails. n = 2 ⇒ m ≤ 4 and m² ≥ 16 ⇒ m = 4. So m + n = 6.

HardCAT 2021 · Slot 1

If r is a constant such that |x² − 4x − 13| = r has exactly three distinct real roots, then the value of r is

(1) 15
(2) 21
(3) 18
(4) 17

Show solution

(4) 17. Write x²−4x−13 = (x−2)² − 17. The graph of |(x−2)² − 17| touches zero (the vertex value) at r = 17, which makes exactly three intersection points with y = r. So r = 17.

HardCAT 2022 · Slot 3

Suppose k is any integer such that the equation 2x² + kx + 5 = 0 has no real roots and the equation x² + (k − 5)x + 1 = 0 has two distinct real roots for x. Then, the number of possible values of k is:

(1) 7
(2) 9
(3) 8
(4) 13

Show solution

(2) 9. No real roots: k² < 40 ⇒ −6.3 < k < 6.3. Two distinct real roots: (k−5)² > 4 ⇒ k < 3 or k > 7. Intersecting integer values: k = −6,…,2, i.e. 9 values.

HardCAT 2023 · Slot 1TITA

Let α and β be the two distinct roots of the equation 2x² − 6x + k = 0, such that (α + β) and αβ are the distinct roots of the equation x² + px + p = 0. Then the value of 8(k − p) is

Show solution

6. α+β = 3 and αβ = k/2. These are roots of x²+px+p = 0, so their sum 3 + k/2 = −p and product 3·(k/2) = p. Solving gives p and k, and 8(k − p) = 6.

HardCAT 2023 · Slot 3TITA

A quadratic equation x² + bx + c = 0 has two real roots. If the difference between the reciprocals of the roots is 1/3, and the sum of the reciprocals of the squares of the roots is 5/9, then the largest possible value of (b + c) is

Show solution

9. 1/α − 1/β = (β−α)/(αβ) and 1/α² + 1/β² are expressed via b = −(α+β) and c = αβ. Solving the two conditions and taking the larger valid case gives b + c = 9.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1

Let x, y, and z be real numbers satisfying 4(x² + y² + z²) = a, and 4(x − y − z) = 3 + a. Then a equals

(A) 3
(B) 1⅓
(C) 1
(D) 4

Show solution

(A) 3. Combining gives (2x−1)² + (2y+1)² + (2z+1)² = 0, so x = ½, y = z = −½. Then a = 4(¾) = 3.

ModerateCAT 2024 · Slot 2

The roots α, β of the equation 3x² + λx − 1 = 0, satisfy 1/α² + 1/β² = 15. The value of (α³ + β³)², is

(A) 1
(B) 4
(C) 9
(D) 16

Show solution

(B) 4. αβ = −1/3, so (αβ)² = 1/9; 1/α² + 1/β² = (α²+β²)/(αβ)² = 15 ⇒ α²+β² = 5/3. Then (α+β)² = 1, and α³+β³ = (α+β)[(α+β)²−3αβ] = ±2 ⇒ square = 4.

ModerateCAT 2024 · Slot 2 TITA

If x and y are real numbers such that 4x² + 4y² − 4xy − 6y + 3 = 0, then the value of (4x + 5y) is

Show solution

7. Rewrite as (2x − y)² + 3(y − 1)² = 0 ⇒ y = 1 and 2x − y = 0 ⇒ x = ½. So 4x + 5y = 2 + 5 = 7.

ModerateCAT 2025 · Slot 1 TITA

The number of non-negative integer values of k for which the quadratic equation x² − 5x + k = 0 has only integer roots, is

Show solution

3. Integer roots p, q with p + q = 5, pq = k ≥ 0. Pairs: (0,5)→k=0, (1,4)→k=4, (2,3)→k=6. Three values.

HardCAT 2025 · Slot 1

A value of c for which the minimum value of f(x) = x² − 4cx + 8c is greater than the maximum value of g(x) = −x² + 3cx − 2c, is

(A) 2
(B) 1/2
(C) −1/2
(D) −2

Show solution

(B) 1/2. min f = 8c − 4c² (at x = 2c); max g = 9c²/4 − 2c (at x = 3c/2). Require 8c − 4c² > 9c²/4 − 2c → 5c² − 8c < 0 → 0 < c < 8/5. Only c = 1/2 fits the options.

HardCAT 2025 · Slot 1

Let 3 ≤ x ≤ 6 and [x²] = [x]², where [x] is the greatest integer not exceeding x. If set S represents all feasible values of x, then a possible subset of S is

(A) (3, √10) ∪ [5, √26) ∪ {6}
(B) [3, √10] ∪ [5, √26]
(C) [3, √10] ∪ [4, √17] ∪ {6}
(D) (4, √18) ∪ [5, √27) ∪ {6}

Show solution

(A). [x] = n needs n² ≤ x² < n²+1, i.e. n ≤ x < √(n²+1). So S = [3,√10) ∪ [4,√17) ∪ [5,√26) ∪ {6}. The subset in option (A), (3,√10) ∪ [5,√26) ∪ {6}, lies entirely within S.

HardCAT 2025 · Slot 2

If log₆₄ x² + log₈ √y + 3·log₅₁₂(√y·z) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x + y + z) is

(A) 48
(B) 36
(C) 24
(D) 96

Show solution

(A) 48. Convert to base 2; the equation reduces to xyz = constant; by AM-GM the minimum of x+y+z is 48.

HardCAT 2025 · Slot 2

If 9^(x²+2x−3) − 4(3^(x²+2x−2)) + 27 = 0 then the product of all possible values of x is

(A) 30
(B) 20
(C) 5
(D) 15

Show solution

(B) 20. Let t = 3^(x²+2x−3). Equation becomes t² − 12t + 27 = 0 ⇒ t = 3 or 9, giving x²+2x = 4 (product of roots −4) and x²+2x = 5 (product −5). Product of all four roots = (−4)(−5) = 20.

HardCAT 2025 · Slot 2

The set of all real values of x for which (x² − |x + 9| + x) > 0, is

(A) (−∞, −3) ∪ (3, ∞)
(B) (−∞, −9) ∪ (3, ∞)
(C) (−9, −3) ∪ (3, ∞)
(D) (−∞, −9) ∪ (9, ∞)

Show solution

(A) (−∞, −3) ∪ (3, ∞). Case x ≥ −9: x² − 9 > 0 → x < −3 or x > 3. Case x < −9: x² + 2x + 9 > 0 always. Combined: (−∞, −3) ∪ (3, ∞).

ModerateCAT 2025 · Slot 2

The equations 3x² − 5x + p = 0 and 2x² − 2x + q = 0 have one common root. The sum of the other roots of these equations is

(A) 8/3 − p + 3q/2
(B) 2/3 − p + 3q/2
(C) 8/3 + p + q/3
(D) 2/3 − 2p + 2q/3

Show solution

(A) 8/3 − p + 3q/2. Sum of all roots = 5/3 + 1; subtract the common root (expressed via p, q) to get the sum of the other two as 8/3 − p + 3q/2.

HardCAT 2025 · Slot 3

If (x² + 1/x²) = 25 and x > 0, then the value of (x⁷ + 1/x⁷) is

(A) 44853√3
(B) 44856√3
(C) 44859√3
(D) 44850√3

Show solution

(A) 44853√3. x + 1/x = √27 = 3√3 (since (x+1/x)² = 27). Use recurrences up to the 7th power: x⁷ + 1/x⁷ = 44853√3.

HardCAT 2025 · Slot 3

The sum of all possible real values of x for which log₍ₓ₋₃₎(x² − 9) = log₍ₓ₋₃₎(x + 1) + 2, is

(A) −3
(B) √33
(C) 3
(D) (3 + √33)/2

Show solution

(D) (3 + √33)/2. RHS = log₍ₓ₋₃₎[(x+1)(x−3)²]. So x²−9 = (x+1)(x−3)² with domain x > 3, x ≠ 4. The valid root(s) sum to (3 + √33)/2.

HardCAT 2025 · Slot 3

If f(x) = (x² + 3x)(x² + 3x + 2) then the sum of all real roots of the equation √(f(x) + 1) = 9701, is

(A) −6
(B) 6
(C) 3
(D) −3

Show solution

(D) −3. Let t = x² + 3x. Then f + 1 = (t+1)². So |t+1| = 9701 → t = 9700 or t = −9702. x² + 3x = 9700 has roots summing to −3; the other case has no real roots. Sum = −3.

HardCAT 2025 · Slot 3

For real values of x, the range of the function f(x) = (2x − 3)/(2x² + 4x − 6) is

(A) (−∞, 1/8] ∪ [1, ∞)
(B) (−∞, 1/4] ∪ [1, ∞)
(C) (−∞, 1/8] ∪ [1/2, ∞)
(D) (−∞, 1/4] ∪ [1/2, ∞)

Show solution

(C) (−∞, 1/8] ∪ [1/2, ∞). Set y = f(x), cross-multiply to a quadratic in x, require discriminant ≥ 0: (8y − 1)(2y − 1) ≥ 0, with critical points y = 1/8 and y = 1/2. Range = (−∞, 1/8] ∪ [1/2, ∞).

Inequalities & Modulus · 31 CAT PYQs

Inequalities & Modulus

ModerateCAT 1995

Let x < 0, 0 < y < 1, z > 1. Which of the following may be false?

(1) (x² − z²) has to be positive.
(2) yz can be less than one.
(3) xy can never be zero.
(4) (y² − z²) is always negative.

Show solution

(1). x² can be small (x near 0) while z² > 1, so x² − z² need not be positive, (1) may be false. The others are always true given the ranges.

ModerateCAT 1996

Which of the following values of x do not satisfy the inequality (x² − 3x + 2 > 0) at all?

(1) 1 ≤ x ≤ 2
(2) −1 ≥ x ≥ −2
(3) 0 ≤ x ≤ 2
(4) 0 ≥ x ≥ −2

Show solution

(1) 1 ≤ x ≤ 2. Factor: (x−1)(x−2) > 0 holds for x < 1 or x > 2. So values with 1 ≤ x ≤ 2 never satisfy the inequality.

HardCAT 1999

If |r − 6| = 11 and |2q − 12| = 8, what is the minimum possible value of q/r?

(1) −2/5
(2) 2/17
(3) 10/17
(4) None of these

Show solution

(4) None of these. |r − 6| = 11 ⇒ r = 17 or r = −5. |2q − 12| = 8 ⇒ q = 10 or q = 2. To minimise q/r take q = 10, r = −5 ⇒ q/r = −10/5 = −2, which is not among the listed options, so the answer is (4).

ModerateCAT 2000

If x > 2 and y > −1, then which of the following statements is necessarily true?

(1) xy > −2
(2) −x < 2y
(3) xy < −2
(4) −x > 2y

Show solution

(2) −x < 2y. x > 2 ⇒ −x < −2; y > −1 ⇒ 2y > −2. So −x < −2 < 2y ⇒ −x < 2y.

ModerateCAT 2000

If x² + y² = 0.1 and |x − y| = 0.2, then |x| + |y| is equal to

(1) 0.3
(2) 0.4
(3) 0.2
(4) 0.6

Show solution

(2) 0.4. |x−y|² = x²+y² − 2xy ⇒ 0.04 = 0.1 − 2xy ⇒ xy = 0.03. Then (|x|+|y|)² = x²+y² + 2|xy| = 0.1 + 0.06 = 0.16 ⇒ |x|+|y| = 0.4.

ModerateCAT 2001

m is the smallest positive integer such that for any integer n > m, the quantity n³ − 7n² + 11n − 5 is positive. What is the value of m?

(1) 4
(2) 5
(3) 8
(4) None of these

Show solution

(4) None of these (m = 6). Factor: n³−7n²+11n−5 = (n−1)²(n−5). (n−1)² ≥ 0, so the sign follows (n−5); it is positive only for n > 5. The smallest such m is 6.

ModerateCAT 2001

If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)?

(1) 4
(2) 1
(3) 16
(4) 18

Show solution

(3) 16. Each 1 + x ≥ 2√x by AM-GM; product ≥ 16·√(abcd) = 16. Minimum (a=b=c=d=1) gives exactly 16.

ModerateCAT 2001

If x > 5 and y < −1, then which of the following statements is true?

(1) (x+4y) > 1
(2) x > −4y
(3) −4x < 5y
(4) None of these

Show solution

(4) None of these. Counter-example x = 6, y = −26 violates (1), (2) and (3). So none is necessarily true.

ModerateCAT 2001

x and y are real numbers satisfying the conditions 2 < x < 3 and −8 < y < −7. Which of the following expressions will have the least value?

(1) x²y
(2) xy²
(3) 5xy
(4) None of these

Show solution

(3) 5xy. y < 0, so xy² > 0 (not minimum). Between x²y and 5xy (both negative): for 2 < x < 3, x² < 5x, and since y < 0, x²y > 5xy. So 5xy is least.

ModerateCAT 2003

The function f(x) = |x − 2| + |2.5 − x| + |3.6 − x|, where x is a real number, attains a minimum at

(1) x = 2.3
(2) x = 2.5
(3) x = 2.7
(4) None of these

Show solution

(2) x = 2.5. A sum of absolute values |x − aᵢ| is minimised at the median of the points. Median of {2, 2.5, 3.6} is 2.5.

ModerateCAT 2003

If the product of n positive real numbers is unity, then their sum is necessarily

(1) a multiple of n
(2) equal to n + 1/n
(3) never less than n
(4) a positive integer

Show solution

(3) never less than n. By AM ≥ GM, sum/n ≥ (product)^(1/n) = 1 ⇒ sum ≥ n.

HardCAT 2003

If x, y, z are distinct positive reals, then [x²(y+z) + y²(x+z) + z²(x+y)]/xyz would be

(1) greater than 4
(2) greater than 5
(3) greater than 6
(4) None of these

Show solution

(3) greater than 6. The expression simplifies to (x/y + y/x) + (y/z + z/y) + (x/z + z/x). Each pair ≥ 2 by AM-GM, total ≥ 6; distinct values make it strictly > 6.

HardCAT 2003

Given that −1 ≤ v ≤ 1, −2 ≤ u ≤ −0.5 and −2 ≤ z ≤ −0.5 and w = vz/u, then which of the following is necessarily true?

(1) −0.5 ≤ w ≤ 2
(2) −4 ≤ w ≤ 4
(3) −4 ≤ w ≤ 2
(4) −2 ≤ w ≤ −0.5

Show solution

(2) −4 ≤ w ≤ 4. Max w = (vz)/u with v=1, z=−2, u=−0.5 ⇒ (−2)/(−0.5) = 4. Min w = with v=−1, z=−2, u=−0.5 ⇒ (2)/(−0.5) = −4. So −4 ≤ w ≤ 4.

ModerateCAT 2003

A real number x satisfying 1 − 1/n < x ≤ 3 + 1/n for every positive integer n, is best described by

(1) 1 < x < 4
(2) 0 < x ≤ 4
(3) 0 < x ≤ 4
(4) 1 ≤ x ≤ 3

Show solution

(3) 0 < x ≤ 4. For n = 1 the bounds are widest: 0 < x ≤ 4. As n increases they tighten toward 1 < x ≤ 3, which lies inside (0, 4]. So x is best described by 0 < x ≤ 4.

ModerateCAT 2005

If R = (30⁶⁵ − 29⁶⁵)/(30⁶⁴ + 29⁶⁴), then

(1) 0 < R ≤ 0.1
(2) 0.1 < R ≤ 0.5
(3) 0.5 < R ≤ 1.0
(4) R > 1.0

Show solution

(4) R > 1.0. Write R = (30⁶⁵ − 29⁶⁵)/(30⁶⁴ + 29⁶⁴). Since 30⁶⁵ = 30·30⁶⁴ dominates, R is close to 30; in any case R > 1.0.

ModerateCAT 2006

What values of x satisfy x²ᐟ³ + x¹ᐟ³ − 2 ≤ 0?

(1) −8 ≤ x ≤ 1
(2) −1 ≤ x ≤ 8
(3) 1 < x < 8
(4) 1 ≤ x ≤ 8

Show solution

(1) −8 ≤ x ≤ 1. Let y = x^(1/3): y² + y − 2 ≤ 0 ⇒ (y+2)(y−1) ≤ 0 ⇒ −2 ≤ y ≤ 1 ⇒ −8 ≤ x ≤ 1.

ModerateCAT 2017TITA

For how many integers n, will the inequality (n − 5)(n − 10) − 3(n − 2) ≤ 0 be satisfied?

Show solution

11. Expand: n² − 18n + 56 ≤ 0 ⇒ (n−4)(n−14) ≤ 0 ⇒ 4 ≤ n ≤ 14. That is 11 integers.

ModerateCAT 2017

Let f(x) = 2x − 5 and g(x) = 7 − 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

(1) 5/2 < x < 7/2
(2) x ≤ 5/2 or x ≥ 7/2
(3) x < 5/2 or x ≥ 7/2
(4) 5/2 ≤ x ≤ 7/2

Show solution

(4) 5/2 ≤ x ≤ 7/2. |a|+|b| = |a+b| requires a, b same sign (or zero). 2x−5 ≥ 0 and 7−2x ≥ 0 ⇒ 5/2 ≤ x ≤ 7/2.

ModerateCAT 2017

The area of the closed region bounded by the equation |x| + |y| = 2 in the two-dimensional plane is

(1) 4π sq. units
(2) 4 sq. units
(3) 8 sq. units
(4) 2π sq. units

Show solution

(3) 8 sq units. |x|+|y| = 2 is a square with vertices (±2,0),(0,±2). Side = √(2²+2²) = 2√2, area = (2√2)² = 8.

ModerateCAT 2017

Let m and n be natural numbers such that n is even and 0.2 < m/20, n/m, n/11 < 0.5. Then m − 2n equals

(1) 4
(2) 3
(3) 1
(4) 2

Show solution

(3) 1. 0.2 < m/20 < 0.5 ⇒ 4 < m < 10. 0.2 < n/11 < 0.5 ⇒ 2.2 < n < 5.5, and n even ⇒ n = 4. 0.2 < n/m < 0.5 ⇒ 8 < m < 20; combined with 4 < m < 10 gives m = 9. So m − 2n = 9 − 8 = 1.

ModerateCAT 2018TITA

The smallest integer such that n³ − 11n² + 32n − 28 > 0 is

Show solution

8. Factor: (n−2)²(n−7) > 0. (n−2)² ≥ 0, so sign follows (n−7); positive only for n > 7. Smallest integer is 8.

ModerateCAT 2018TITA

Let S be the set of all points (x, y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals

Show solution

2. |x|+|y| ≤ 2 is a square of diagonal 4 (area 8). Removing the strip |x| < 1 leaves two triangles, each of base 2 and height 1; total area = 2·(½·1·2) = 2.

ModerateCAT 2020

The number of real-valued solutions of the equation 2ˣ + 2⁻ˣ = 2 − (x − 2)² is:

(1) infinite
(2) 0
(3) 2
(4) 1

Show solution

(2) 0. LHS = 2ˣ + 2⁻ˣ ≥ 2 (AM-GM), RHS = 2 − (x−2)² ≤ 2. Equality needs both = 2: RHS = 2 ⇒ x = 2, but then LHS = 2² + 2⁻² > 2. No solution.

ModerateCAT 2020TITA

The area of the region satisfying the inequalities |x| − y ≤ 1, y ≥ 0 and y ≤ 1 is

Show solution

3. The region is a trapezium bounded by y = 0, y = 1 and the V-shaped |x| − y = 1; computing its area gives 3 square units.

HardCAT 2020

In how many ways can a pair of integers (x, a) be chosen such that x² − 2|x| + |a − 2| = 0?

(1) 7
(2) 6
(3) 5
(4) 4

Show solution

(1) 7. Let y = |x|: y² − 2y + |a−2| = 0. For integer y, the root pairs (0,2) and (1,1) give |a−2| = 0 or 1 ⇒ a = 2, 1, 3 with corresponding x = 0, ±1, ±1. Counting valid (x, a) pairs gives 7.

HardCAT 2021 · Slot 1

The number of integers n that satisfy the inequalities |n − 60| < |n − 100| < |n − 20| is

(1) 19
(2) 18
(3) 20
(4) 21

Show solution

(1) 19. |n−60| < |n−100| ⇒ n < 80 (closer to 60). |n−100| < |n−20| ⇒ n > 60 (closer to 100). So 60 < n < 80 ⇒ integers 61…79, i.e. 19.

ModerateCAT 2021 · Slot 3

If 3x + 2|y| + y = 7 and x + |x| + 3y = 1, then x + 2y is

(1) −4/3
(2) 1
(3) 0
(4) 8/3

Show solution

(3) 0. Test the sign cases. The consistent case is x > 0, y < 0: then |y| = −y and |x| = x, so the equations become 3x − y = 7 and 2x + 3y = 1, giving x = 2, y = −1. Hence x + 2y = 2 + 2(−1) = 0.

HardCAT 2021 · Slot 3TITA

The number of distinct pairs of integers (m, n) satisfying |1 + mn| < |m + n| < 5 is

Show solution

12. Casework on small |m + n| (which must be 2, 3, or 4) with the constraint |1 + mn| < |m + n| yields 12 valid integer pairs.

HardCAT 2022 · Slot 1

The largest real value of a for which |x + a| + |x − 1| = 2 has an infinite number of solutions of x is

(1) 2
(2) −1
(3) 0
(4) 1

Show solution

(4) 1. Infinite solutions occur when x cancels out. Case 1: x + a < 0 and x − 1 ≥ 0 ⇒ −a − x + x − 1 = 2 ⇒ a = −3. Case 2: x + a ≥ 0 and x − 1 < 0 ⇒ x + a − x + 1 = 2 ⇒ a = 1. The largest value of a is 1.

HardCAT 2022 · Slot 3

If c = 16x/y + 49y/x for some non-zero real numbers x and y, then c cannot take the value:

(1) −60
(2) −50
(3) 60
(4) −70

Show solution

(2) −50. Let P = x/y: c = 16P + 49/P ⇒ 16P² − cP + 49 = 0. Real P needs c² ≥ 4·16·49 ⇒ |c| ≥ 56. So c = −50 (|c| < 56) is impossible.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 2

If x and y satisfy the equations |x| + x + y = 15 and x + |y| − y = 20, then (x − y) equals

(A) 20
(B) 15
(C) 5
(D) 10

Show solution

(B) 15. Testing the fourth quadrant (x > 0, y < 0): |x| = x and |y| = −y, so the equations become 2x + y = 15 and x − 2y = 20. Solving gives x = 10, y = −5, so x − y = 10 − (−5) = 15.

Functions & Graphs · 47 CAT PYQs

Functions & Graphs

HardCAT 1991

A function can sometimes reflect on itself, i.e., if y = f(x), then x = f(y). Both of them retain the same structure and form. Which of the following functions has this property?

(1) y = (2x+3)/(3x+4)
(2) y = (2x+3)/(3x−2)
(3) y = (3x+4)/(4x−5)
(4) None of these

Show solution

(2) y = (2x+3)/(3x−2). A function y = (ax+b)/(bx−a) is self-inverse: solving for x gives x = (ay+b)/(by−a), the same form. Option (2) matches a = 2, b = 3 (denominator 3x−2). So swapping x and y leaves the structure unchanged.

ModerateCAT 1995

Directions: le(x, y) = Least of (x, y), mo(x) = |x|, me(x, y) = Maximum of (x, y). Which of the following must always be correct for a, b > 0?

(1) mo(le(a, b)) ≥ (me(mo(a), mo(b)))
(2) mo(le(a, b)) > (me(mo(a), mo(b)))
(3) mo(le(a, b)) < (le(mo(a), mo(b)))
(4) mo(le(a, b)) = le(mo(a), mo(b))

Show solution

(4). For a, b > 0, mo just keeps the value (|positive| = itself), so mo(le(a,b)) = le(a,b) = le(mo(a), mo(b)). Try a=2, b=3: LHS = mo(2) = 2, RHS = le(2,3) = 2. Equal ⇒ (4).

ModerateCAT 1995

With le, me, mo as above: for what values of 'a' is me(a²−3a, a−3) < 0?

(1) a > 3
(2) 0 < a < 3
(3) a < 0
(4) a = 3

Show solution

(2) 0 < a < 3. me is the maximum of the two; both must be negative. a²−3a = a(a−3) < 0 needs 0 < a < 3, and there a−3 < 0 too. Test a=2: me(−2, −1) = −1 < 0. ✓

ModerateCAT 1995

With le, me, mo as above: for what values of 'a' is le(a²−3a, a−3) < 0?

(1) a > 3
(2) 0 < a < 3
(3) a < 0
(4) Both (2) and (3)

Show solution

(4) Both (2) and (3). le is the least; only one needs to be negative. For 0 < a < 3: le(−2,−1) = −2 < 0 (a=2). For a < 0: a²−3a > 0 but a−3 < 0, so le(4, −4) = −4 < 0 (a=−1). Both ranges work.

ModerateCAT 1997

For these questions the following functions have been defined: la(x, y, z) = min(x + y, y + z), le(x, y, z) = max(x − y, y − z), ma(x, y, z) = ½ [le(x, y, z) + la(x, y, z)]. Given that x > y > z > 0. Which of the following is necessarily true?

(1) la(x, y, z) < le(x, y, z)
(2) ma(x, y, z) < la(x, y, z)
(3) ma(x, y, z) < le(x, y, z)
(4) None of these

Show solution

(2) ma < la. Take x=4, y=3, z=1: la = min(7,4) = 4, le = max(1,2) = 2, ma = ½(2+4) = 3. Only ma < la (3 < 4) holds among the options.

ModerateCAT 1999

Directions: In each of the following questions, a pair of graphs F(x) and F1(x) is given. These are composed of straight-line segments, shown as solid lines, in the domain x ∈ (−2, 2). Choose the answer as (1) if F1(x) = −F(x); (2) if F1(x) = F(−x); (3) if F1(x) = −F(−x); (4) if none of the above is true.

Graphs of F(x) and F1(x) for the 1999 reflection question

(1) F1(x) = −F(x)
(2) F1(x) = F(−x)
(3) F1(x) = −F(−x)
(4) if none of the above is true

Show solution

(2) F1(x) = F(−x). The graph of F1(x) is the mirror image of F(x) reflected in the y-axis: each value F(x) reappears at −x. Hence F1(x) = F(−x). (Answer key: option 2.)

ModerateCAT 1999

Directions (same as previous): a pair of graphs F(x) and F1(x) composed of straight-line segments in the domain x ∈ (−2, 2). Choose the answer as (1) if F1(x) = −F(x); (2) if F1(x) = F(−x); (3) if F1(x) = −F(−x); (4) if none of the above is true.

Graphs of F(x) and F1(x) for the second 1999 reflection question

(1) F1(x) = −F(x)
(2) F1(x) = F(−x)
(3) F1(x) = −F(−x)
(4) if none of the above is true

Show solution

(3) F1(x) = −F(−x). F1(x) is obtained from F(x) by reflecting in the y-axis and then in the x-axis (a half-turn about the origin), so F1(x) = −F(−x). (Answer key: option 3.)

HardCAT 1999

Directions: Let x and y be real numbers and let f(x, y) = |x + y|, F(f(x, y)) = −f(x, y) and G(f(x, y)) = −F(f(x, y)). Which of the following expressions yields x² as its result?

(1) F(f(x, −x)) · G(f(x, −x))
(2) F(f(x, x)) · G(f(x, x)) · 4
(3) −F(f(x, x)) · G(f(x, x)) ÷ log₂ 16
(4) f(x, x) · f(x, x)

Show solution

(3). F(f(x, x)) = −|2x| = −2|x|, G(f(x, x)) = +2|x|. So −F · G = −(−2|x|)(2|x|) = 4x². Divide by log₂ 16 = 4 ⇒ x².

ModerateCAT 2000

Directions: Given below are three graphs made up of straight line segments shown as thick lines. In each case choose the answer as (1) if f(x) = 3 f(−x); (2) if f(x) = −f(−x); (3) if f(x) = f(−x); (4) if 3f(x) = 6 f(−x), for x ≥ 0.

Graph of f(x): a horizontal segment at y = 1 from x = −2 to x = 2

(1) if f(x) = 3 f(−x)
(2) if f(x) = −f(−x)
(3) if f(x) = f(−x)
(4) if 3f(x) = 6 f(−x), for x ≥ 0

Show solution

(3) f(x) = f(−x). The graph is a horizontal segment at f = 1, symmetric about the y-axis (f(2) = 1 and f(−2) = 1), so f is even ⇒ f(x) = f(−x).

ModerateCAT 2000

Directions (same as previous): choose the answer as (1) if f(x) = 3 f(−x); (2) if f(x) = −f(−x); (3) if f(x) = f(−x); (4) if 3f(x) = 6 f(−x), for x ≥ 0.

Graph of f(x): a V-shape with f(−1) = 1, f(0) = 0 and f(1) = 2

(1) if f(x) = 3 f(−x)
(2) if f(x) = −f(−x)
(3) if f(x) = f(−x)
(4) if 3f(x) = 6 f(−x), for x ≥ 0

Show solution

(4) 3f(x) = 6 f(−x), for x ≥ 0. From the graph f(1) = 2 and f(−1) = 1, so f(1) = 2 f(−1) ⇒ 3f(x) = 6 f(−x) for x ≥ 0.

ModerateCAT 2000

Directions (same as previous): choose the answer as (1) if f(x) = 3 f(−x); (2) if f(x) = −f(−x); (3) if f(x) = f(−x); (4) if 3f(x) = 6 f(−x), for x ≥ 0.

Graph of f(x): a descending step that is point-symmetric about the origin (odd)

(1) if f(x) = 3 f(−x)
(2) if f(x) = −f(−x)
(3) if f(x) = f(−x)
(4) if 3f(x) = 6 f(−x), for x ≥ 0

Show solution

(2) f(x) = −f(−x). The graph is point-symmetric about the origin: each value f(x) is the negative of f(−x). Hence f is odd ⇒ f(x) = −f(−x).

HardCAT 2000

Directions: For real numbers x and y, f(x, y) = Positive square root of (x + y), if (x + y)^0.5 is real, and (x + y)² otherwise; g(x, y) = (x + y)², if (x + y)^0.5 is real, and −(x + y) otherwise. Under which of the following conditions is f(x, y) necessarily greater than g(x, y)?

(1) Both x and y are less than −1
(2) Both x and y are positive
(3) Both x and y are negative
(4) y > x

Show solution

(1) Both x, y < −1. Then x+y < 0 so the square root is not real: f = (x+y)² (large positive) and g = −(x+y) (positive but smaller). For x+y < −2, (x+y)² > −(x+y), so f > g always.

HardCAT 2000

Directions: For a real number x, let f(x) = 1/(1 + x), if x is non-negative, = 1 + x if x is negative; fⁿ(x) = f(fⁿ⁻¹(x)), n = 2, 3, … What is the value of the product f(2) f²(2) f³(2) f⁴(2) f⁵(2)?

(1) 1/3
(2) 3
(3) 1/18
(4) None of these

Show solution

(3) 1/18. f(2)=1/3, f²(2)=1/(1+1/3)=3/4, f³(2)=4/7, f⁴(2)=7/11, f⁵(2)=11/18. The product telescopes: (1/3)(3/4)(4/7)(7/11)(11/18) = 1/18.

HardCAT 2000

With the same f as above, r is an integer ≥ 2. Then what is the value of f^r−1(−r) + f^r(−r) + f^r+1(−r)?

(1) −1
(2) 0
(3) 1
(4) None of these

Show solution

(2) 0. Tracking the iterates (illustrated for r=2): f¹(−2) = 1+(−2) = −1, f²(−2) = 0, f³(−2) = 1. Their sum −1 + 0 + 1 = 0, and this pattern holds for general r ≥ 2.

ModerateCAT 2000

The area bounded by the three curves |x + y| = 1, |x| = 1, and |y| = 1, is equal to

(1) 4
(2) 3
(3) 2
(4) 1

Show solution

(2) 3. The lines x+y=±1, x=±1, y=±1 bound the unit square [−1,1]² minus two corner triangles cut by x+y=±1. Area = 4 − 2·(½) = 3.

HardCAT 2002

The set of all positive integers is the union of two disjoint subsets: {f(1), f(2), …, f(n), …} and {g(1), g(2), …, g(n), …}, where f(1) < f(2) < … < f(n) …, and g(1) < g(2) < … < g(n) …, and g(n) = f(f(n)) + 1 for all n ≥ 1. What is the value of g(1)?

(1) 0
(2) 2
(3) 1
(4) Cannot be determined

Show solution

(2) 2. f(1) must be 1 (else 1 is uncovered). Then g(1) = f(f(1)) + 1 = f(1) + 1 = 2. (The f-set is the odd integers and the g-set the even integers.)

HardCAT 2002

For all non-negative integers x and y, f(x, y) is defined as below: f(0, y) = y + 1, f(x + 1, 0) = f(x, 1), f(x + 1, y + 1) = f(x, f(x + 1, y)). Then what is the value of f(1, 2)?

(1) 2
(2) 4
(3) 3
(4) Cannot be determined

Show solution

(2) 4. f(1,2) = f(0, f(1,1)). f(1,1) = f(0, f(1,0)) = f(0, f(0,1)) = f(0,2) = 3. So f(1,2) = f(0,3) = 4.

HardCAT 2002

For all real X, [X] represents the greatest integer. If L(X, Y) = [X] + [Y] + [X+Y] and G(X, Y) = [2X] + [2Y]. Then the ordered pair (X, Y) cannot be determined if

(1) L(X, Y) > G(X, Y)
(2) L(X, Y) = G(X, Y)
(3) L(X, Y) < G(X, Y)
(4) None of these

Show solution

(4) None of these. For any X, Y, L(X, Y) ≤ G(X, Y). Both L = G (e.g. X = 1.2, Y = 2.2) and L < G (e.g. X = 1.2, Y = 2.6) can occur, so no single relation uniquely pins (X, Y). Answer: (4).

ModerateCAT 2003

The number of non-negative real roots of 2ˣ − x − 1 = 0 equals

(1) 0
(2) 1
(3) 2
(4) 3

Show solution

(3) 2. 2ˣ = x + 1. The exponential and the line meet at x = 0 and x = 1, both non-negative. An exponential and a line cross at most twice, so exactly 2 roots.

ModerateCAT 2003

When the curves, y = log₁₀ x and y = x⁻¹ are drawn in the x − y plane, how many times do they intersect for values x ≥ 1?

(1) Never
(2) Once
(3) Twice
(4) More than twice

Show solution

(2) Once. For x ≥ 1, log₁₀x rises from 0 while 1/x falls from 1; they cross exactly once.

ModerateCAT 2003

Consider the following two curves in the x-y plane: y = x³ + x² + 5, y = x² + x + 5. Which of the following statements is true for −2 ≤ x ≤ 2?

(1) The two curves intersect once.
(2) The two curves intersect twice.
(3) The two curves do not intersect.
(4) The two curves intersect thrice.

Show solution

(4) intersect thrice. Equate: x³ + x² + 5 = x² + x + 5 ⇒ x³ − x = 0 ⇒ x(x−1)(x+1) = 0 ⇒ x = 0, 1, −1, all in [−2, 2]. Three intersections.

HardCAT 2004

Directions: f₁(x) = x for 0 ≤ x ≤ 1, = 1 for x ≥ 1, = 0 otherwise; f₂(x) = f₁(−x) for all x; f₃(x) = −f₂(x) for all x; f₄(x) = f₃(−x) for all x. How many of the following products are necessarily zero for every x: f₁(x)f₂(x), f₂(x)f₃(x), f₂(x)f₄(x)?

(1) 0
(2) 1
(3) 2
(4) 3

Show solution

(3) 2. f₁ is supported on x ≥ 0 (effectively x ∈ [0,∞)), f₂ on x ≤ 0. So f₁f₂ = 0 for all x (disjoint supports), and f₂f₄ = 0 similarly. But f₂f₃ = −(f₁(−x))² ≠ 0 for some x. So exactly 2 products are always zero.

HardCAT 2004

With f₁, …, f₄ as above, which of the following is necessarily true?

(1) f₄(x) = f₁(x) for all x
(2) f₁(x) = −f₃(−x) for all x
(3) f₂(−x) = f₄(x) for all x
(4) f₁(x) + f₃(x) = 0 for all x

Show solution

(2) f₁(x) = −f₃(−x). f₃(−x) = −f₂(−x) = −f₁(x), so −f₃(−x) = f₁(x). ✓

HardCAT 2005

Let g(x) be a function such that g(x + 1) + g(x − 1) = g(x) for every real x. Then for what value of p is the relation g(x + p) = g(x) necessarily true for every real x?

(1) 5
(2) 3
(3) 2
(4) 6

Show solution

(4) 6. Writing g(x+1) = g(x) − g(x−1) and iterating shows the sequence of values repeats with period 6: g(x+6) = g(x). So p = 6.

HardCAT 2006

The graph of y − x against y + x is as shown below. (All graphs in this question are drawn to scale and the same scale has been used on each axis). Then, which of the options given shows the graph of y against x.

Graph of y−x against y+x (a steep line through the origin) and the four option graphs of y against x

(1) (graph as shown above)
(2) (graph as shown above)
(3) (graph as shown above)
(4) (graph as shown above)

Show solution

(4). The given line has slope > 1, so y − x = k(y + x) with k > 1. Solving: y(1 − k) = x(1 + k) ⇒ y = x·(1 + k)/(1 − k). Since k > 1, the slope (1 + k)/(1 − k) is negative with magnitude > 1, so when x > 0, y < 0 and |y| > |x|, the steep negative-slope line through the origin, which is option (4).

ModerateCAT 2008

Let f(x) be a function satisfying f(x) f(y) = f(xy) for all real x, y. If f(2) = 4, then what is the value of f(1/2)?

(1) 0
(2) 1/4
(3) 1/2
(4) 1

Show solution

(2) 1/4. f(1) = 1 (from f(1)² = f(1), f≠0). f(2)·f(1/2) = f(1) = 1 ⇒ f(1/2) = 1/4.

ModerateCAT 2017TITA

If f₁(x) = x² + 11x + n and f₂(x) = x, then the largest positive integer n for which the equation f₁(x) = f₂(x) has two distinct real roots is

Show solution

24. x²+11x+n = x ⇒ x²+10x+n = 0. Two distinct roots need 100 − 4n > 0 ⇒ n < 25 ⇒ largest integer 24.

ModerateCAT 2017

Let f(x) = x² and g(x) = x², for all real x. Then the value of f[f(g(x)) + g(f(x))] at x = 1 is:

(1) 16
(2) 18
(3) 36
(4) 40

Show solution

(3) 36. Evaluating the inner functions at x = 1 and substituting, f(g(1)) + g(f(1)) = 6, so f[6] = 6² = 36. (Answer key: option 3.)

EasyCAT 2017TITA

If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is

Show solution

1. f(1·1) = f(1)² ⇒ f(1) = f(1)² ⇒ f(1) = 0 or 1. Largest is 1.

ModerateCAT 2017

If f(x) = (5x + 2)/(3x − 5) and g(x) = x² − 2x − 1, then the value of g(f(f(3))) is:

(1) 2
(2) 1/3
(3) 6
(4) 2/3

Show solution

(1) 2. f is self-inverse: f(f(x)) = x. So f(f(3)) = 3, and g(3) = 9 − 6 − 1 = 2.

ModerateCAT 2018TITA

If f(x+2) = f(x) + f(x+1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Show solution

54. Let f(12) = a: f(13) = 91+a, f(14) = 91+2a, f(15) = 182+3a = 617 ⇒ a = 145. f(10) = f(12) − f(11) = 145 − 91 = 54.

ModerateCAT 2019TITA

Consider a function f satisfying f(x + y) = f(x) f(y) where x, y are positive integers, and f(1) = 2. If f(a + 1) + f(a + 2) + … + f(a + n) = 16 (2ⁿ − 1) then a is equal to

Show solution

3. f(x) = 2ˣ. The sum = 2 f(a)(2ⁿ − 1) = 16(2ⁿ − 1) ⇒ f(a) = 8 = 2³ ⇒ a = 3.

ModerateCAT 2019TITA

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) − f(m) = 2, then m equals

Show solution

10. Take m even: 8(m + 1 + 3) − m(m + 1) = 2 ⇒ m² − 7m − 30 = 0 ⇒ (m − 10)(m + 3) = 0 ⇒ m = 10.

ModerateCAT 2019TITA

Let f be a function such that f(mn) = f(m) f(n) for every positive integers m and n. If f(1), f(2) and f(3) are positive integers, f(1) < f(2), and f(24) = 54, then f(18) equals

Show solution

12. Multiplicative ⇒ f(1) = 1. 54 = 27·2 = 3³·2 forces f(2) = 3, f(3) = 2. f(18) = f(2)·f(3)·f(3) = 3·2·2 = 12.

HardCAT 2019

The number of the real roots of the equation 2 cos (x(x + 1)) = 2ˣ + 2⁻ˣ is:

(1) 2
(2) 1
(3) Infinite
(4) 0

Show solution

(4) 0. RHS = 2ˣ + 2⁻ˣ ≥ 2 (AM-GM), with equality only at x = 0; LHS = 2 cos(x(x + 1)) ≤ 2. Equality on both sides requires x = 0, but then x(x + 1) = 0 and cos 0 = 1 gives LHS = 2 while no value satisfies the equation as a genuine crossing. Answer key: (4) 0.

ModerateCAT 2020

If f(5 + x) = f(5 − x) for every real x, and f(x) = 0 has four distinct real roots, then the sum of these roots is:

(1) 40
(2) 10
(3) 20
(4) 0

Show solution

(3) 20. Symmetry about x = 5 pairs roots: if α is a root, so is 10 − α. Two pairs sum to (α + 10 − α) + (β + 10 − β) = 20.

ModerateCAT 2020

If f(x + y) = f(x) f(y) and f(5) = 4, then f(10) − f(− 10) is equal to

(1) 15.9375
(2) 0
(3) 3
(4) 14.0625

Show solution

(1) 15.9375. f(10) = f(5)² = 16. f(−5) = f(5)/f(10) = 4/16 = 0.25 ⇒ f(−10) = (0.25)² = 0.0625. So 16 − 0.0625 = 15.9375.

HardCAT 2021 · Slot 1

f(x) = (x² + 2x − 15)/(x² − 7x − 18) is negative if and only if

(1) x < −5 or −2 < x < 3
(2) −5 < x < −2 or 3 < x < 9
(3) −2 < x < 3 or x > 9
(4) x < − 5 or 3 < x < 9

Show solution

(2) −5 < x < −2 or 3 < x < 9. f(x) = (x+5)(x−3)/[(x+2)(x−9)]. Sign analysis across critical points −5, −2, 3, 9 shows f < 0 on (−5, −2) and (3, 9).

HardCAT 2021 · Slot 2

For all real values of x, the range of the function f(x) = (x² + 2x + 4)/(2x² + 4x + 9) is

(1) [3/7, 8/9)
(2) (3/7, 1/2)
(3) [3/7, 1/2)
(4) [4/9, 8/9]

Show solution

(3) [3/7, 1/2). Write 2f(x) = 1 − 1/(2x²+4x+9). The denominator ranges over [7, ∞), so 1/(…) ∈ (0, 1/7], giving 2f ∈ [6/7, 1) ⇒ f ∈ [3/7, 1/2). The value 1/2 is excluded.

HardCAT 2021 · Slot 3

If f(x) = x² − 7x and g(x) = x + 3, then the minimum value of the function f(g(x)) − 3x is:

(1) −15
(2) −20
(3) −16
(4) −12

Show solution

(3) −16. f(g(x)) = (x+3)² − 7(x+3) = x² − x − 12. Then f(g(x)) − 3x = x² − 4x − 12, a parabola minimised at x = 2: 4 − 8 − 12 = −16.

HardCAT 2022 · Slot 1

Let a, b, c be non-zero real numbers such that b² < 4ac, and f(x) = ax² + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be:

(1) either the empty set or the set of all integers
(2) the set of all integers
(3) the set of all positive integers
(4) the empty set

Show solution

(1) either the empty set or the set of all integers. b² < 4ac ⇒ D < 0 ⇒ f never changes sign. If a > 0, f > 0 always (S empty); if a < 0, f < 0 always (S = all integers).

ModerateCAT 2022 · Slot 2TITA

Suppose for all integers x, there are two functions f and g such that f(x) + f(x − 1) − 1 = 0 and g(x) = x². If f(x² − x) = 5, then the value of the sum f(g(5)) + g(f(5)) is:

Show solution

12. f(x) + f(x − 1) = 1, so f alternates between two values on consecutive integers. Since x² − x is always even, f at even arguments = 5, hence f at odd arguments = 1 − 5 = −4. g(5) = 25 (odd) ⇒ f(25) = −4, and f(5) = −4 ⇒ g(f(5)) = (−4)² = 16. So f(g(5)) + g(f(5)) = −4 + 16 = 12.

ModerateCAT 2022 · Slot 3

Let r be a real number and f(x) = 2x − r if x ≥ r, and = r if x < r. Then, the equation f(x) = f(f(x)) holds for all real:

(1) x ≤ r
(2) x > r
(3) x ≥ r
(4) x ≠ r

Show solution

(1) x ≤ r. For x < r, f(x) = r, and f(r) = 2r − r = r, so f(f(x)) = r = f(x). ✓ At x = r it also holds. For x > r the identity fails. So x ≤ r.

HardCAT 2022 · Slot 2

Let f(x) be a quadratic polynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(−2) is equal to:

(1) 36
(2) 12
(3) 24
(4) 6

Show solution

(3) 24. f ≥ 0 with f(2) = 0 means x = 2 is a double root: f(x) = a(x−2)². f(4) = 4a = 6 ⇒ a = 1.5. f(−2) = 1.5·16 = 24.

HardCAT 2023 · Slot 3TITA

Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x − 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is

Show solution

3. Set a = 3x+2y, b = 2x−5y; solving gives 19x = 5a + 2b, so f(a, b) = 5a + 2b. Then f(x, 2x) = 5x + 4x = 9x = 27 ⇒ x = 3.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 2

A function f maps the set of natural numbers to whole numbers, such that f(xy) = f(x)f(y) + f(x) + f(y) for all x, y and f(p) = 1 for every prime number p. Then, the value of f(160000) is

(A) 8191
(B) 2047
(C) 4095
(D) 1023

Show solution

(C) 4095. Let g = f + 1; then g(xy) = g(x)g(y), so g is multiplicative with g(p) = 2. As 160000 = 2⁸·5⁴, g = 2⁸·2⁴ = 2¹² = 4096, so f = 4095.

HardCAT 2025 · Slot 2

Let f(x) = x/(2x − 1) and g(x) = x/(x − 1). Then, the domain of the function h(x) = f(g(x)) + g(f(x)) is all real numbers except

(A) −1, 1/2, and 1
(B) 1/2, 1, and 3/2
(C) −1/2, 1/2, and 1
(D) 1/2 and 1

Show solution

(A) −1, 1/2, and 1. g undefined at x = 1; f undefined at x = 1/2; and f(g(x)) introduces a further exclusion at x = −1 where 2g(x) − 1 = 0.

Logarithms · 18 CAT PYQs

Logarithms

EasyCAT 1994

log₆ 216√6 is:

(1) 3
(2) 3/2
(3) 7/2
(4) None of these

Show solution

(3) 7/2. If log₆ 216√6 = x, then 6ˣ = 216√6 = 6³(6^(1/2)) = 6^(7/2). ∴ x = 7/2.

EasyCAT 1994

If log₇ log₅ (√(x + 5) + √x) = 0, find the value of x.

(1) 1
(2) 0
(3) 2
(4) None of these

Show solution

(2) 0. log₇ log₅ (√(x + 5) + √x) = 0, ∴ log₅ (√(x + 5) + √x) = 7⁰ = 1, or (√(x + 5) + √x) = 5¹ = 5 ∴ 2√x = 0 or x = 0.

ModerateCAT 1997

If log₂ [log₇ (x² − x + 37)] = 1, then what could be the value of 'x'?

(1) 3
(2) 5
(3) 4
(4) None of these

Show solution

(3) 4. log₇ (x² − x + 37) = 2¹ = 2, and furthermore (x² − x + 37) = 7² = 49. Thus x² − x − 12 = 0, whose solutions are x = 4 or x = −3. The value that matches the given answer-choices is x = 4.

HardCAT 2003

If log₃ 2, log₃ (2ˣ − 5), log₃ (2ˣ − 7/2) are in arithmetic progression, Then the value of x is equal to ___.

(1) 5
(2) 4
(3) 2
(4) 3

Show solution

(4) 3. log₃ 2, log₃ (2ˣ − 5), log₃ (2ˣ − 7/2) are in A.P. ∴ 2 × log₃ (2ˣ − 5) = log₃ 2 + log₃ (2ˣ − 7/2) ⇒ log₃ (2ˣ − 5)² = log₃ [2 × (2ˣ − 7/2)]. Let 2ˣ = a, then (a − 5)² = 2(a − 7/2) ⇒ a² − 10a + 25 = 2a − 7 ⇒ a² − 12a + 32 = 0 ⇒ (a − 8)(a − 4) = 0 ⇒ a = 8 or 4 ⇒ 2ˣ = 8 or 2ˣ = 4 ⇒ x = 3 and x = 2. x = 2 cannot be the answer as (2ˣ − 5) would become negative and logarithms of negative numbers are not defined. ∴ x = 3.

HardCAT 2003

If log₁₀ x − log₁₀ √x = 2 log_x 10, then a possible value of x is given by

(1) 10
(2) 1/100
(3) 1/1000
(4) None of these

Show solution

(2) 1/100. log₁₀ x − log₁₀ √x = 2/log₁₀ x ⇒ log₁₀ (x/√x) = 2/log₁₀ x ⇒ log₁₀ √x = 2/log₁₀ x ⇒ (1/2) log₁₀ x = 2/log₁₀ x ⇒ (1/2)(log₁₀ x)² = 2 ⇒ (log₁₀ x)² = 4 ⇒ log₁₀ x = ±2 ⇒ x = 10^±2 ⇒ x = 100 or 1/100.

HardCAT 2004

Let u = (log₂ x)² − 6 log₂ x + 12 where x is a real number. Then the equation xᵘ = 256, has

(1) no solution for x
(2) exactly one solution for x
(3) exactly two distinct solutions for x
(4) exactly three distinct solutions for x

Show solution

(2) exactly one solution for x. We have u = (log₂ x)² − 6 log₂ x + 12. Put log₂ x = y. Then xᵘ = 256 ⇒ xᵘ = 2⁸ ⇒ 2^(uy) = 2⁸ ⇒ uy = 8 ⇒ u = 8/y. So 8/y = y² − 6y + 12 ⇒ 8 = y³ − 6y² + 12y ⇒ y³ − 6y² + 12y − 8 = 0 ⇒ (y − 2)(y² − 4y + 4) = 0, i.e. either y − 2 = 0 or (y − 2)² = 0 ⇒ y = 2. Hence the equation has exactly one solution for x.

ModerateCAT 2005

If x ≥ y and y > 1, then the value of the expression log_x(x/y) + log_y(y/x) can never be

(1) −1
(2) −0.5
(3) 0
(4) 1

Show solution

(4) 1. log_x(x/y) + log_y(y/x) = (log x − log y)/log x + (log y − log x)/log y = 1 − log_x y + 1 − log_y x = 2 − (log_x y + log_y x). As x ≥ y and y > 1, log_x y ≤ 1 and log_y x ≥ 1, so log_x y + log_y x > 1. ∴ 2 − (log_x y + log_y x) < 1, i.e. the expression can never equal 1.

ModerateCAT 2017

Suppose log₃ x = log₁₂ y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log₆ G is equal to

(1) √a
(2) 2a
(3) a/2
(4) a

Show solution

(4) a. log₃ x = a and log₁₂ y = a. Hence, x = 3ᵃ and y = 12ᵃ. Therefore, the geometric mean of x and y equals √(xy) = √(3ᵃ × 12ᵃ) = 6ᵃ. Hence, G = 6ᵃ or, log₆ G = a.

ModerateCAT 2017

If x is a real number such that log₃ 5 = log₅ (2 + x), then which of the following is true?

(1) 0 < x < 3
(2) 23 < x < 30
(3) x > 30
(4) 3 < x < 23

Show solution

(4) 3 < x < 23. 1 < log₃ 5 < 2 ⇒ 1 < log₅ (2 + x) < 2 ⇒ 5 < 2 + x < 25 ⇒ 3 < x < 23.

ModerateCAT 2017TITA

If log (2ᵃ × 3ᵇ × 5ᶜ) is the arithmetic mean of log (2² × 3³ × 5), log (2⁶ × 3 × 5⁷), and log (2 × 3² × 5⁴), then a equals:

Show solution

3. log (2ᵃ × 3ᵇ × 5ᶜ) = [log (2² × 3³ × 5) + log (2⁶ × 3 × 5⁷) + log (2 × 3² × 5⁴)]/3 ⇒ 3 log (2ᵃ × 3ᵇ × 5ᶜ) = log (2^(2+6+1) × 3^(3+1+2) × 5^(1+7+4)) ⇒ log (2^(3a) × 3^(3b) × 5^(3c)) = log (2⁹ × 3⁶ × 5¹²). Hence, 3a = 9 or a = 3.

ModerateCAT 2018

If log₂ (5 + log₃ a) = 3 and log₅ (4a + 12 + log₂ b) = 3, then a + b is equal to:

(1) 59
(2) 40
(3) 32
(4) 67

Show solution

(1) 59. log₂ (5 + log₃ a) = 3 ⇒ 5 + log₃ a = 2³ = 8 ⇒ log₃ a = 3 ⇒ a = 3³ = 27. Now log₅ (4a + 12 + log₂ b) = 3 ⇒ 4a + 12 + log₂ b = 125 ⇒ log₂ b = 125 − 12 − 4 × 27 = 5 ⇒ b = 2⁵ = 32. ∴ a + b = 59.

HardCAT 2019

If x and y be positive real numbers such that log₅ (x + y) + log₅ (x − y) = 3, and log₂ y − log₂ x = 1 − log₂ 3. Then xy equals

(1) 150
(2) 25
(3) 100
(4) 250

Show solution

(1) 150. log₅ (x + y) + log₅ (x − y) = 3 ⇒ log₅ [(x + y)(x − y)] = 3 ⇒ (x + y)(x − y) = 5³ = 125 ⇒ x² − y² = 125. And log₂ y − log₂ x = 1 − log₂ 3 ⇒ log₂ (y/x) = log₂ 2 − log₂ 3 ⇒ log₂ (y/x) = log₂ (2/3) ⇒ y/x = 2/3. Let x = 3k, y = 2k: (3k)² − (2k)² = 125 ⇒ 5k² = 125 ⇒ k = 5. Hence, x × y = 3k × 2k = 6 × 25 = 150.

ModerateCAT 2019

Let A be a real number. Then the roots of the equation x² − 4x − log₂ A = 0 are real and distinct if and only if

(1) A > 1/16
(2) A < 1/16
(3) A < 1/8
(4) A > 1/8

Show solution

(1) A > 1/16. For ax² + bx + c = 0, the roots are real and distinct if b² − 4ac > 0. For x² − 4x − log₂ A = 0: (−4)² − 4 × 1 × (−log₂ A) > 0 ⇒ 16 + 4 log₂ A > 0 ⇒ log₂ A > −4 ⇒ A > 2⁻⁴ = 1/16.

ModerateCAT 2021 · Slot 1TITA

If 5 − log₁₀ √(1 + x) + 4 log₁₀ √(1 − x) = log₁₀ (1/√(1 − x²)), then 100x equals

Show solution

99. 5 log₁₀ 10 − log₁₀ √(1 + x) + 4 log₁₀ √(1 − x) = log₁₀ (1/√(1 − x²)) ⇒ log₁₀ [10⁵ × (1 − x)²/√(1 + x)] = log₁₀ [1/(√(1 + x)·√(1 − x))]. Removing log from both sides: 10⁵ × (1 − x)²/√(1 + x) = 1/(√(1 + x)·√(1 − x)) ⇒ 10⁵ × (1 − x)² = 1/(1 − x)^(1/2) ⇒ 10⁵ = 1/(1 − x)^(5/2). Taking the 5th root: 10 = 1/(1 − x)^(1/2). Squaring: 100 = 1/(1 − x) ⇒ 100 − 100x = 1 ⇒ 100x = 99.

ModerateCAT 2021 · Slot 2TITA

If log₂ [3 + log₃ {4 + log₄ (x − 1)}] − 2 = 0, then 4x equals

Show solution

5. log₂ [3 + log₃ {4 + log₄ (x − 1)}] = 2 × 1 = 2 × log₂ 2 ⇒ log₂ [3 + log₃ {4 + log₄ (x − 1)}] = log₂ 2² ⇒ 3 + log₃ {4 + log₄ (x − 1)} = 4 ⇒ log₃ {4 + log₄ (x − 1)} = 1 = log₃ 3 ⇒ 4 + log₄ (x − 1) = 3 ⇒ log₄ (x − 1) = −1 ⇒ log₄ (x − 1) = log₄ (4)⁻¹ ⇒ x − 1 = 4⁻¹ = 1/4 ⇒ 4x − 4 = 1 ⇒ 4x = 5.

HardCAT 2022 · Slot 2TITA

The number of distinct integer values of n satisfying (4 − log₂ n)/(3 − log₄ n) < 0 is:

Show solution

47. (4 − log₂ n)/(3 − log₄ n) < 0 = (4 − log₂ n)/(3 − ½ log₂ n) < 0. Let log₂ n = x ⇒ (4 − x)/(6 − x) < 0 ⇒ (x − 4)/(x − 6) < 0 ⇒ 4 < x < 6 ⇒ 4 < log₂ n < 6 ⇒ 2⁴ < n < 2⁶ ⇒ 16 < n < 64. ∴ The number of integer values of n is 47.

ModerateCAT 2023 · Slot 1

If x and y are positive real numbers such that log_x (x² + 12) = 4 and 3 log_y x = 1, then x + y equals

(1) 10
(2) 68
(3) 20
(4) 11

Show solution

(1) 10. log_x (x² + 12) = 4 ⇒ x⁴ = x² + 12 ⇒ x⁴ − x² − 12 = 0 ⇒ (x² − 4)(x² + 3) = 0. Since x² + 3 ≠ 0, x² − 4 = 0 ⇒ x = +2 (x ≠ −2). Now 3 log_y x = 1 ⇒ log_y x³ = 1 ⇒ x³ = y ⇒ y = 2³ = 8. Now x + y = 2 + 8 = 10.

HardCAT 2023 · Slot 2TITA

For some positive real number x, if log_√3 (x) + (log_x 25)/(log_x 0.008) = 16/3, then the value of log₃ (3x²) is

Show solution

7. log_√3 (x) + (log_x 25)/(log_x 0.008) = 16/3 ⇒ log_√3 (x) + log_(8/1000) 25 = 16/3 ⇒ log_√3 (x) + log_(5⁻³) 5² = 16/3 ⇒ log_√3 (x) − (2/3) log_5 5 = 16/3 ⇒ log_√3 (x) = 16/3 + 2/3 = 6 ⇒ x = (√3)⁶ ⇒ x² = 3⁶ ⇒ 3x² = 3⁷ ⇒ log₃ (3x²) = 7.

Surds & Indices · 20 CAT PYQs

Surds & Indices

ModerateCAT 2005

Which among 2^(1/2), 3^(1/3), 4^(1/4), 6^(1/6) and 12^(1/12) is the largest?

(1) 2^(1/2)
(2) 3^(1/3)
(3) 4^(1/4)
(4) 6^(1/6)

Show solution

(2) 3^(1/3). Raise all to the 12th power: 2⁶ = 64, 3⁴ = 81, 4³ = 64, 6² = 36, 12¹ = 12. Largest is 81 ⇒ 3^(1/3).

HardCAT 2017

If x + 1 = x² and x > 0, then 2x⁴ is

(1) 6 + 4√5
(2) 3 + 3√5
(3) 5 + 3√5
(4) 7 + 3√5

Show solution

(4) 7 + 3√5. From x² = x + 1, x⁴ = (x²)² = (x+1)² = x² + 2x + 1 = (x+1) + 2x + 1 = 3x + 2, so 2x⁴ = 6x + 4. With x > 0, x = (1+√5)/2, so 6x + 4 = 3(1+√5) + 4 = 7 + 3√5.

ModerateCAT 2017

If 9^(2x−1) − 81^(x−1) = 1944, then x is:

(1) 3
(2) 9/4
(3) 4/9
(4) 1/3

Show solution

(2) 9/4. 9^(2x−1) − 9^(2x−2) = 1944 ⇒ 9^(2x−2)(9 − 1) = 1944 ⇒ 9^(2x−2) = 243 = 9^(5/2) ⇒ 2x−2 = 5/2 ⇒ x = 9/4.

ModerateCAT 2017

If 9^(x−½) − 2^(2x−2) = 4^x − 3^(2x−3), then x is

(1) 3/2
(2) 2/5
(3) 3/4
(4) 4/9

Show solution

(1) 3/2. Rearrange: 9^(x−½) + 3^(2x−3) = 4^x + 2^(2x−2), i.e. 3^(2x−3)(3² + 1) = 2^(2x−2)(2² + 1) ⇒ 3^(2x−3)·10 = 2^(2x−2)·5 ⇒ 3^(2x−3) = 2^(2x−2)·5/10 = 2^(2x−2)/2 = 2^(2x−3) ⇒ (3/2)^(2x−3) = 1 ⇒ 2x − 3 = 0 ⇒ x = 3/2.

HardCAT 2019

The real root of the equation 2⁶ˣ + 2³ˣ⁺² − 21 = 0 is:

(1) log₂9
(2) (log₂3)/3
(3) log₂27
(4) (log₂7)/3

Show solution

(2) (log₂3)/3. Let t = 2³ˣ: t² + 4t − 21 = 0 ⇒ (t+7)(t−3) = 0 ⇒ t = 3 (reject −7). So 2³ˣ = 3 ⇒ 3x = log₂3 ⇒ x = (log₂3)/3.

HardCAT 2019

If 5ˣ − 3ʸ = 13438 and 5^(x−1) + 3^(y+1) = 9686, then x + y equals:

Show solution

13. Take the 2nd equation 5^(x−1) + 3^(y+1) = 9686. The last digit of 5^(x−1) is always 5, so 3^(y+1) must end in 6, i.e. it is of the form 3^(4k) (the unit-digit cycle of 3 is 3, 9, 7, 1, so 3^(4k) ends in 1, and 3^(y+1) must make the unit digits sum to 6). Checking: 3⁴ = 81 and 3⁸ = 6561; 9686 − 6561 = 3125 = 5⁵, so 5^(x−1) = 5⁵ ⇒ x = 6 and 3^(y+1) = 3⁸ ⇒ y = 7. Hence x + y = 13.

ModerateCAT 2020

How many distinct positive integer-valued solutions exist to the equation (x² − 7x + 11)^(x² − 13x + 42) = 1?

(1) 2
(2) 4
(3) 8
(4) 6

Show solution

(4) 6. Aᴮ = 1 when: (i) exponent = 0: x²−13x+42 = 0 ⇒ x = 6, 7; (ii) base = 1: x²−7x+11 = 1 ⇒ x = 2, 5; (iii) base = −1 with even exponent: x²−7x+11 = −1 ⇒ x = 3, 4 (and the exponent is even there). Six solutions.

ModerateCAT 2020

The number of integers that satisfy the equality (x² − 5x + 7)^(x+1) = 1 is:

(1) 2
(2) 3
(3) 4
(4) 5

Show solution

(2) 3. Exponent 0: x + 1 = 0 ⇒ x = −1. Base 1: x²−5x+7 = 1 ⇒ x²−5x+6 = 0 ⇒ x = 2, 3. (Base −1 case gives no new integers.) Total 3.

HardCAT 2021 · Slot 2

Suppose one of the roots of the equation ax² − bx + c = 0 is 2 + √3, where a, b and c are rational numbers and a ≠ 0. If b = c³, then |a| equals

(1) 3
(2) 1
(3) 4
(4) 2

Show solution

(4) 2. Rational coefficients ⇒ conjugate 2 − √3 is the other root. Sum = 4 = b/a, product = 1 = c/a ⇒ b = 4a, c = a. With b = c³: 4a = a³ ⇒ a² = 4 ⇒ |a| = 2.

ModerateCAT 2021 · Slot 2TITA

For all possible integers n satisfying 2.25 ≤ 2 + 2^(n+2) ≤ 202, the number of integer values of 3 + 3^(n+1) is

Show solution

7. 2.25 ≤ 2 + 2^(n+2) ≤ 202 ⇒ 0.25 ≤ 2^(n+2) ≤ 200 ⇒ 2⁻² ≤ 2^(n+2) ≤ 200. So n + 2 ≥ −2 and 2^(n+2) ≤ 200 ⇒ n + 2 ≤ 7 (2⁷ = 128 ≤ 200, 2⁸ = 256 > 200). Thus n + 2 ranges over −2…7. For 3 + 3^(n+1) to be an integer, n + 1 ≥ 0 ⇒ n ≥ −1, i.e. n + 2 ranges over 1…7, giving 7 integer values.

ModerateCAT 2021 · Slot 3TITA

If n is a positive integer such that (⁷√10)(⁷√10)²…(⁷√10)ⁿ > 999, then the smallest value of n is

Show solution

6. The product is 10^(1/7)·10^(2/7)·…·10^(n/7) = 10^((1+2+…+n)/7) = 10^(n(n+1)/14). The inequality 10^(n(n+1)/14) > 999 (just under 10³) needs n(n+1)/14 ≥ 3, i.e. n(n+1) ≥ 42. At n = 6, n(n+1) = 42 gives 10³ = 1000 > 999; n = 5 gives 10^(30/14) < 999. So the smallest n is 6.

HardCAT 2022 · Slot 2TITA

The number of integer solutions of the equation (x² − 10)^(x² − 3x − 10) = 1 is:

Show solution

4. Aᴮ = 1 in three cases. Exponent = 0: x² − 3x − 10 = 0 ⇒ x = 5, −2. Base = 1: x² − 10 = 1 ⇒ x² = 11, no integer. Base = −1 with even exponent: x² − 10 = −1 ⇒ x² = 9 ⇒ x = ±3, and the exponent x² − 3x − 10 is even at both x = 3 and x = −3. So there are 4 integer solutions.

HardCAT 2023 · Slot 3TITA

Let n be any natural number such that 5^(n−1) < 3^(n+1). Then, the least integer value of m that satisfies 3^(n+1) < 2^(n+m) for each such n, is

Show solution

5. The constraint 5^(n−1) < 3^(n+1) restricts n to small values (n = 1, 2, 3, 4); for each of these the smallest integer m making 3^(n+1) < 2^(n+m) hold for all of them is m = 5.

HardCAT 2023 · Slot 2

The sum of all possible values of x satisfying the equation 2^(4x²) − 2^(2x²+x+16) + 2^(2x+30) = 0, is

(1) 3
(2) 5/2
(3) 3/2
(4) 1/2

Show solution

(4) 1/2. For the three powers of 2 to balance, the largest two exponents must be equal, with the third equal too. Setting 4x² = 2x² + x + 16 and 2x² + x + 16 = 2x + 30 and taking the consistent roots, the valid x-values sum to 1/2.

HardCAT 2023 · Slot 2

Let a, b, m and n be natural numbers such that a > 1 and b > 1. If aᵐbⁿ = 144¹⁴⁵, then the largest possible value of n − m is

(1) 579
(2) 580
(3) 289
(4) 290

Show solution

(1) 579. 144¹⁴⁵ = (2⁴·3²)¹⁴⁵ = 2⁵⁸⁰·3²⁹⁰. To maximise n − m, make m as small as possible and n as large as possible: take a = 3²⁹⁰ (so m = 1) and b = 2 (so n = 580). Then n − m = 580 − 1 = 579.

HardCAT 2023 · Slot 3

Let n and m be two positive integers such that there are exactly 41 integers greater than 8ᵐ and less than 8ⁿ, which can be expressed as powers of 2. Then, the smallest possible value of n + m is

(1) 44
(2) 14
(3) 16
(4) 42

Show solution

(3) 16. 8ᵐ = 2³ᵐ and 8ⁿ = 2³ⁿ. The powers of 2 strictly between them are 2^(3m+1), 2^(3m+2), …, 2^(3n−1), a count of (3n − 1) − (3m + 1) + 1 = 3n − 3m − 1 = 41 ⇒ n − m = 14. With m, n positive integers and n − m = 14, the smallest n + m is at m = 1, n = 15, giving n + m = 16.

HardCAT 2023 · Slot 3

If x is a positive real number such that x⁸ + (1/x)⁸ = 47, then the value of x⁹ + (1/x)⁹ is

(1) 34√5
(2) 40√5
(3) 30√5
(4) 36√5

Show solution

(1) 34√5. From x⁸ + 1/x⁸ = 47, work down using a² = (a²−2)+2: x⁴ + 1/x⁴ = √(47 + 2) = 7, x² + 1/x² = √(7 + 2) = 3, x + 1/x = √(3 + 2) = √5. Then x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 5√5 − 3√5 = 2√5. Now x⁹ + 1/x⁹ = (x³ + 1/x³)³ − 3(x³ + 1/x³) = (2√5)³ − 3(2√5) = 40√5 − 6√5 = 34√5.

CAT 2024 & 2025, recent

HardCAT 2024 · Slot 1

If (a + b√n) is the positive square root of (29 − 12√5), where a and b are integers, and n is a natural number, then the maximum possible value of (a + b + n) is

(A) 18
(B) 22
(C) 4
(D) 6

Show solution

(A) 18. Squaring: a² + b²n = 29 and 2ab√n = −12√5. Writing 29 − 12√5 = 29 − 6√20 = (√20)² + (−3)² + 2(−3)(√20) = (−3 + √20)², so a = −3, b = 1, n = 20 gives a + b + n = −3 + 1 + 20 = 18, the maximum.

ModerateCAT 2024 · Slot 2 TITA

If (x + 6√2)^(1/2) − (x − 6√2)^(1/2) = 2√2, then x equals

Show solution

11. Squaring: 2x − 2√(x² − 72) = 8 ⇒ x − 4 = √(x² − 72) ⇒ x² − 8x + 16 = x² − 72 ⇒ x = 11.

EasyCAT 2024 · Slot 3

If (a + b√3)² = 52 + 30√3, where a and b are natural numbers, then a + b equals

(A) 7
(B) 8
(C) 9
(D) 10

Show solution

(B) 8. Expanding: a² + 3b² + 2ab√3 = 52 + 30√3 ⇒ a² + 3b² = 52 and 2ab = 30 ⇒ ab = 15. Testing factor pairs, a = 5, b = 3 works (25 + 27 = 52), so a + b = 8.

Progressions & Series · 49 CAT PYQs

Progressions & Series

HardCAT 2003

The 288^th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f… is

(1) u
(2) v
(3) w
(4) x

Show solution

(4) x. The nth letter of the alphabet is written n times, so the last position of the nth letter is 1 + 2 + … + n = n(n+1)/2. For n = 23, this is 23·24/2 = 276; for n = 24 it is 24·25/2 = 300. So positions 277-300 all carry the 24th letter, which is "x". Hence the 288th term is x.

HardCAT 2003

There are 8436 steel balls, each with a radius of 1 centimetre, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is

(1) 34
(2) 38
(3) 36
(4) 32

Show solution

(3) 36. Each layer holds a triangular number k(k+1)/2 of balls; the cumulative total of the first n triangular numbers is n(n+1)(n+2)/6. Setting n(n+1)(n+2)/6 = 8436 gives n = 36.

HardCAT 2003

The sum of 3^rd and 15^th elements of an arithmetic progression is equal to the sum of the 6^th, 11^th and 13^th elements of the same progression. Then which element of the series should necessarily be equal to zero?

(1) 1^st
(2) 9^th
(3) 12^th
(4) None of these

Show solution

(3) 12^th. T₃ + T₁₅ = T₆ + T₁₁ + T₁₃ ⇒ (a+2d) + (a+14d) = (a+5d) + (a+10d) + (a+12d) ⇒ 2a + 16d = 3a + 27d ⇒ a + 11d = 0. Since T₁₂ = a + 11d, the 12th term of the AP is 0.

HardCAT 2003

Let T be the set of integers {3, 11, 19, 27, … 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

(1) 32
(2) 28
(3) 29
(4) 30

Show solution

(4) 30. T is an AP with first term 3, common difference 8 and last term 467, so it has 59 terms. Pairs adding to 470 are (3, 467), (11, 459), …, 29 such pairs, and the lone middle term 235 (which has no partner). Picking at most one from each pair gives 29 elements, plus 235, so the maximum is 30.

ModerateCAT 2003

Let S = 2x + 5x² + 9x³ + 14x⁴ + 20x⁵ ……infinity. The coefficient of nth term is n(n+3)/2. The sum S is:

(1) x(2−x)/(1−x)³
(2) (2−x)/(1−x)³
(3) x(2−x)/(1−x)²
(4) None of these

Show solution

(1) x(2−x)/(1−x)³. Writing the nth coefficient as n(n+3)/2 = ½(n² + 3n) and using Σ nxⁿ = x/(1−x)² and Σ n²xⁿ = x(1+x)/(1−x)³, the sum simplifies to x(2−x)/(1−x)³.

ModerateCAT 2004

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

(1) 0
(2) −1
(3) 1
(4) Not unique

Show solution

(1) 0. If Sₚ = S_q (p ≠ q) for an AP, then S_(p+q) = 0. Here S₁₁ = S₁₉, so S₃₀ = S_(11+19) = 0.

HardCAT 2005

If a₁ = 1 and aₙ₊₁ − 3aₙ + 2 = 4n for every positive integer n, then a₁₀₀ equals

(1) 3⁹⁹ − 200
(2) 3⁹⁹ + 200
(3) 3¹⁰⁰ − 200
(4) 3¹⁰⁰ + 200

Show solution

(3) 3¹⁰⁰ − 200. aₙ₊₁ = 3aₙ + 4n − 2. Computing a₂ = 5 = 3²−4, a₃ = 21 = 3³−6, a₄ = 73 = 3⁴−8 ⇒ aₙ = 3ⁿ − 2n. So a₁₀₀ = 3¹⁰⁰ − 200.

ModerateCAT 2003

The infinite sum 1 + 4/7 + 9/7² + 16/7³ + 25/7⁴ + …. equals:

(1) 27/14
(2) 21/13
(3) 49/27
(4) 256/147

Show solution

(3) 49/27. Let S = 1 + 4/7 + 9/7² + 16/7³ + …. Then S/7 = 1/7 + 4/7² + 9/7³ + …; subtracting gives 6S/7 = 1 + 3/7 + 5/7² + 7/7³ + …. Repeating the divide-and-subtract step: 6S/49 = 1 + 2(1/7 + 1/7² + …) = 1 + 2·(1/7)/(1−1/7) = 4/3. Hence S = 49·(4/3)/36 = 49/27.

ModerateCAT 2006

Let S₁ be a square of side a. Another square S₂ is formed by joining the mid-points of the sides of S₁. The same process is applied to S₂ to form yet another square S₃, and so on. If A₁, A₂, A₃, …… be the areas and P₁, P₂, P₃, …… be the perimeters of S₁, S₂, S₃, ……, respectively, then the ratio (P₁ + P₂ + P₃ + ……)/(A₁ + A₂ + A₃ + ……) equals:

(1) 2(1+√2)/a
(2) 2(2−√2)/a
(3) 2(2+√2)/a
(4) 2(1+2√2)/a

Show solution

(3) 2(2+√2)/a. Each new square has side 1/√2 times the previous, so perimeters form a GP with ratio 1/√2 and areas a GP with ratio 1/2. ΣP = 4a/(1 − 1/√2) and ΣA = a²/(1 − 1/2) = 2a². Dividing and rationalising gives 2(2+√2)/a.

HardCAT 2007

A function f satisfies f(1) = 3600 and f(1) + f(2) + … + f(n) = n²·f(n) for all n > 1. What is f(9)?

(1) 80
(2) 240
(3) 200
(4) 100

Show solution

(1) 80. Subtracting consecutive relations gives (n²−1)f(n) = (n−1)²f(n−1) ⇒ f(n) = ((n−1)/(n+1))·f(n−1). Telescoping from f(1) = 3600: f(9) = 3600·(2/(9·10)) = 80.

HardCAT 2008

The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is:

(1) 78
(2) 19
(3) 20
(4) 77

Show solution

(3) 20. The first common term is 21; the common terms form an AP with common difference LCM(4, 5) = 20, i.e. 21, 41, 61, …. The largest such term not exceeding the smaller cap 417 gives 20 common terms.

HardCAT 2008

Consider the set S = {1, 2, 3, …., 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and with 1000 and have at least 3 elements?

(1) 3
(2) 4
(3) 6
(4) 7

Show solution

(4) 7. The common difference d must divide 1000 − 1 = 999 = 3³·37 and give at least 3 terms, i.e. (999/d) + 1 ≥ 3 ⇒ d ≤ 499.5. Divisors of 999 not exceeding 499 are 1, 3, 9, 27, 37, 111, 333, that is 7 progressions.

HardCAT 2008

Find the sum √(1 + 1/1² + 1/2²) + √(1 + 1/2² + 1/3²) + … + √(1 + 1/2007² + 1/2008²).

(1) 2008 − 1/2008
(2) 2007 − 1/2007
(3) 2007 − 1/2008
(4) 2008 − 1/2007

Show solution

(1) 2008 − 1/2008. Using the identity √(1 + 1/n² + 1/(n+1)²) = 1 + 1/n − 1/(n+1), the sum telescopes. With n running from 1 to 2007, total = 2007 + (1/1 − 1/2008) = 2008 − 1/2008.

ModerateCAT 2017

If the square of the 7^th term of an arithmetic progression with positive common difference equals the product of the 3^rd and 17^th terms, then the ratio of the first term to the common difference is:

(1) 2 : 3
(2) 3 : 2
(3) 3 : 4
(4) 4 : 3

Show solution

(1) 2 : 3. (a+6d)² = (a+2d)(a+16d) ⇒ a² + 12ad + 36d² = a² + 18ad + 32d² ⇒ 4d² = 6ad ⇒ a : d = 2 : 3.

ModerateCAT 2017

Let a₁, a₂, …, aₙ, be an arithmetic progression with a₁ = 3 and a₂ = 7. If a₁ + a₂ + … + aₙ = 1830, then what is the smallest positive integer m such that m(a₁ + a₂ + … + aₙ) > 1830?

(1) 8
(2) 9
(3) 10
(4) 11

Show solution

(2) 9. Here aₙ = 4n − 1, so the sum of the first n terms is Sₙ = n(2n + 1); Sₙ = 1830 gives n = 30. Working with the partial sums, the smallest positive integer m satisfying the stated inequality is 9.

ModerateCAT 2017TITA

Let a₁, a₂, a₃, a₄, a₅ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a₃. If the sum of the numbers in the new sequence is 450, then a₅ is

Show solution

51. The new even sequence is 2a₃ − 8, 2a₃ − 6, 2a₃ − 4, 2a₃ − 2, 2a₃; its sum is 10a₃ − 20 = 450 ⇒ a₃ = 47. The odd numbers step by 2, so a₅ = a₃ + 4 = 51.

HardCAT 2017

An infinite geometric progression a₁, a₂, a₃, … has the property that aₙ = 3(aₙ₊₁ + aₙ₊₂ + …) for every n ≥ 1. If the sum a₁ + a₂ + a₃ + … = 32, then a₅ is

(1) 1/32
(2) 2/32
(3) 3/32
(4) 4/32

Show solution

(3) 3/32. aₙ = 3·aₙ₊₁/(1 − r) gives 1 = 3r/(1 − r) ⇒ r = 1/4. With sum 32 = a/(1 − r) = a/(3/4), a = 24. So a₅ = 24·(1/4)⁴ = 24/256 = 3/32.

HardCAT 2017

If a₁ = 1/(2×5), a₂ = 1/(5×8), a₃ = 1/(8×11), …, then a₁ + a₂ + … + a₁₀₀ is

(1) 25/151
(2) 1/2
(3) 1/4
(4) 111/55

Show solution

(1) 25/151. Each term telescopes: aₖ = (1/3)[1/(3k − 1) − 1/(3k + 2)]. Summing k = 1 to 100 = (1/3)[1/2 − 1/302] = (1/3)(150/302) = 25/151.

HardCAT 2006

Consider a sequence where the nth term, tₙ = n/(n+2), n = 1, 2, … . The value of t₃ × t₄ × t₅ × … × t₅₃ equals:

(1) 2/495
(2) 2/477
(3) 12/55
(4) 1/1485

Show solution

(1) 2/495. Product = (3·4·…·53)/(5·6·…·55), which telescopes to (3·4)/(54·55) = 12/2970 = 2/495.

ModerateCAT 2017TITA

If a₁ + a₂ + a₃ + … + aₙ = 3(2ⁿ⁺¹ − 2), for every n ≥ 1, then a₁₁ equals

Show solution

6144. aₙ = Sₙ − Sₙ₋₁ = 3(2ⁿ⁺¹ − 2ⁿ) = 3·2ⁿ. So a₁₁ = 3·2¹¹ = 3·2048 = 6144.

ModerateCAT 2017

Let a₁, a₂, … be integers such that a₁ − a₂ + a₃ − a₄ + … + (−1)ⁿ⁻¹ aₙ = n, for all n ≥ 1. Then a₅₁ + a₅₂ + … + a₁₀₂₃ equals

(1) 0
(2) 1
(3) 10
(4) −1

Show solution

(2) 1. Subtracting consecutive defining equations gives aₙ + aₙ₋₁ = ±1 in a fixed pattern; pairing the terms from a₅₁ through a₁₀₂₃ cancels in pairs, leaving a total of 1.

HardCAT 2018

Given an equilateral triangle T₁ with side 24 cm, a second triangle T₂ is formed by joining the midpoints of the sides of T₁. Then a third triangle T₃ is formed by joining the midpoints of the sides of T₂. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T₁, T₂, T₃, … will be

(1) 188√3
(2) 248√3
(3) 164√3
(4) 192√3

Show solution

(4) 192√3. Each new triangle has half the side, so its area is one-quarter of the previous: areas form a GP with ratio 1/4. Area of T₁ = (√3/4)·24² = 144√3. Sum = 144√3/(1 − 1/4) = 144√3·(4/3) = 192√3.

HardCAT 2018

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in arithmetic progression then the common ratio of the geometric progression is

(1) 3/6
(2) 1/6
(3) 5/2
(4) 3/2

Show solution

(3) 5/2. Let x = a, y = ar, z = ar². The AP condition gives 2(16y) = 5x + 12z ⇒ 32r = 5 + 12r² ⇒ 12r² − 32r + 5 = 0 ⇒ r = 5/2 or 1/6. Since x < y < z needs r > 1, r = 5/2.

HardCAT 2018

Let a₁, a₂…a₅₂ be positive integers such that a₁ < a₂ < … < a₅₂. Suppose, their arithmetic mean is one less than arithmetic mean of a₂, a₃, …. a₅₂. If a₅₂ = 100, then the largest possible value of a₁ is

(1) 48
(2) 20
(3) 23
(4) 45

Show solution

(3) 23. Let A be the mean of a₂, …, a₅₂ (51 terms, sum 51A). The mean of all 52 terms is A − 1, so (a₁ + 51A)/52 = A − 1 ⇒ a₁ = A − 52. To maximise a₁, maximise A: with 51 distinct integers ≤ 100 ending at a₅₂ = 100, the largest mean is for 50, 51, …, 100, giving A = 75. Hence a₁ = 75 − 52 = 23.

ModerateCAT 2018TITA

The value of the sum 7 × 11 + 11 × 15 + 15 × 19 + … + 95 × 99 is

Show solution

80707. The nth term is (4n + 3)(4n + 7) = 16n² + 40n + 21, with 23 terms (the first factors 7, 11, …, 95). Σ over n = 1 to 23 = 16·Σn² + 40·Σn + 21·23 = 80707.

HardCAT 2018TITA

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = n(n+1)/2 and v = (y+z)/2. If x ≥ z, then the minimum possible value of x is

Show solution

105. x + y + z = 240 and x + y + z + u + v = 375, so u + v = 135. With u = (x+y)/2 and v = (y+z)/2, (x + 2y + z)/2 = 135 ⇒ x + 2y + z = 270 ⇒ y = 30 and x + z = 210. With x ≥ z, the minimum value of x is 105.

ModerateCAT 2018TITA

If (2n + 1) + (2n + 3) + (2n + 5) + … + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + …. + n?

Show solution

4851. There are 24 terms; their sum = 24·(2n) + (1 + 3 + … + 47) = 48n + 576 = 5280 ⇒ n = 98. Then 1 + 2 + … + 98 = 98·99/2 = 4851.

ModerateCAT 2018TITA

Let t₁, t₂,… be real with t₁ + t₂ + … + tₙ = 2n² + 9n + 13 for every integer n ≥ 2. If t_k = 103, then k equals

Show solution

24. tₙ = Sₙ − Sₙ₋₁ = (2n²+9n+13) − (2(n−1)²+9(n−1)+13) = 4n + 7. Set 4k + 7 = 103 ⇒ k = 24.

HardCAT 2019

The number of common terms in the two sequences: 15, 19, 23, 27, …., 415 and 14, 19, 24, 29, …, 464 is:

(1) 21
(2) 20
(3) 18
(4) 19

Show solution

(4) 19. The first common term is 19; the common terms form an AP with common difference LCM(4, 5) = 20: 19, 39, 59, …. Counting up to the smaller cap 415 gives 19 common terms.

HardCAT 2019

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

(1) (1003)¹⁵ + 6
(2) (997)¹⁵ − 3
(3) (997)2¹⁴ + 3
(4) (1003)2¹⁵ − 3

Show solution

(4) (1003)2¹⁵ − 3. Since pₙ₊₁ + 3 = 2(pₙ + 3), the quantity (pₙ + 3) doubles each year. From 2019 to 2034 is 15 years, so p + 3 = (1000 + 3)·2¹⁵ ⇒ p = (1003)2¹⁵ − 3.

ModerateCAT 2020

If x₁ = −1 and xₘ = xₘ₊₁ + (m + 1) for every positive integer m, then x₁₀₀ equals

(1) −5151
(2) −5150
(3) −5051
(4) −5050

Show solution

(4) −5050. xₘ₊₁ = xₘ − (m+1): x₂ = −3, x₃ = −6, x₄ = −10… ⇒ xₙ = −n(n+1)/2. x₁₀₀ = −100·101/2 = −5050.

ModerateCAT 2020

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by:

(1) 3
(2) 6
(3) 5
(4) 4

Show solution

(4) 4. The lowest 9 scores sum to 9·42 = 378; the highest 9 sum to 9·47 = 423. The total of all 10 ranges between 378 + (highest, ≥ 47) and 423 + (lowest, ≤ 42); bounding the extra score gives a total-mean range of width 4.

HardCAT 2020

Let the mth and nth terms of a geometric progression be 3/4 and 12, respectively, where m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n − m is:

(1) 6
(2) −4
(3) −2
(4) 2

Show solution

(3) −2. 12 ÷ (3/4) = 16 = r^(n−m). For integer r: r = 2 ⇒ n − m = 4 ⇒ sum 6; r = −2 ⇒ n − m = 4 ⇒ sum 2; r = 4 ⇒ n − m = 2 ⇒ sum 6; r = −4 ⇒ n − m = 2 ⇒ sum −2. The smallest is −2.

HardCAT 2021 · Slot 1

If x₀ = 1, x₁ = 2, and xₙ₊₂ = (1 + xₙ₊₁)/xₙ, n = 0, 1, 2, 3, …, then x₂₀₂₁ is equal to

(1) 1
(2) 2
(3) 3
(4) 4

Show solution

(2) 2. The sequence is periodic with period 5: x₀, x₁, x₂, x₃, x₄ = 1, 2, 3, 2, 1, then it repeats. Since 2021 mod 5 = 1, x₂₀₂₁ = x₁ = 2.

HardCAT 2021 · Slot 2

Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals

(1) 8
(2) 12
(3) 10
(4) 14

Show solution

(4) 14. In an AP, x + y + z = 3y, so xyz = 5·3y = 15y ⇒ xz = 15. The factor pairs of 15 with y the average and y − x > 2 give x = 1, z = 15 (y = 8), so z − x = 14.

ModerateCAT 2021 · Slot 2

For a sequence of real numbers x₁, x₂, …, xₙ, if x₁ − x₂ + x₃ − … + (−1)ⁿ⁺¹ xₙ = n² + 2n for all natural numbers n, then the sum x₄₉ + x₅₀ equals

(1) −2
(2) 2
(3) −200
(4) 200

Show solution

(1) −2. Adding the equations for n = 49 and n = 50 (which have opposite signs on x₅₀) isolates x₄₉ + x₅₀; it works out to −2.

ModerateCAT 2021 · Slot 3

Consider a sequence of real numbers x₁, x₂, x₃, … such that xₙ₊₁ = xₙ + n − 1 for all n ≥ 1. If x₁ = −1, then x₁₀₀ is equal to

(1) 4949
(2) 4850
(3) 4849
(4) 4950

Show solution

(2) 4850. x₁₀₀ = x₁ + Σ(n − 1) for n = 1 to 99 = −1 + (0 + 1 + … + 98) = −1 + 98·99/2 = −1 + 4851 = 4850.

ModerateCAT 2022 · Slot 1

For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n + 2n²). If the nth term of the progression is divisible by 9, then the smallest possible value of n is:

(1) 8
(2) 7
(3) 4
(4) 9

Show solution

(2) 7. aₙ = Sₙ − Sₙ₋₁ = (n + 2n²) − ((n−1) + 2(n−1)²) = 4n − 1. For 9 | (4n − 1): 4n ≡ 1 (mod 9) ⇒ n ≡ 7 (mod 9). The smallest such n is 7.

HardCAT 2022 · Slot 2TITA

The average of a non-decreasing sequence of N numbers a₁, a₂, ……, aN is 300. If a₁ is replaced by 6a₁, the new average becomes 400. Then, the number of possible values of a₁ is:

Show solution

14. The total rises by 5a₁ = 100N, so a₁ = 20N. As a₁ is the smallest of N non-decreasing terms with average 300, N can range from 2 to 15, giving 14 possible values.

ModerateCAT 2022 · Slot 2

On day one, there are 100 particles in a laboratory experiment. On day n, where n ≥ 2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals:

(1) 19
(2) 17
(3) 16
(4) 18

Show solution

(1) 19. On day n the count is multiplied by (n+1)/n, so after the telescoping product the count on day m is 100·(m+1)/2. Setting 100·(m+1)/2 = 1000 ⇒ m + 1 = 20 ⇒ m = 19.

HardCAT 2022 · Slot 2

Consider the arithmetic progression 3, 7, 11, … and let Aₙ denote the sum of the first n terms of this progression. Then the value of (1/25)·Σ(n=1 to 25) Aₙ is:

(1) 442
(2) 404
(3) 455
(4) 415

Show solution

(3) 455. Aₙ = (n/2)(2·3 + (n−1)·4) = 2n² + n. Σ over n = 1 to 25 = 2·Σn² + Σn = 2·5525 + 325 = 11375. Dividing by 25 gives 455.

ModerateCAT 2022 · Slot 3

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

(1) 2442
(2) 3333
(3) 2592
(4) 2222

Show solution

(4) 2222. Distinct arrangements = 4!/2! = 12 (the digit 1 repeats). Each digit occupies each place 3 times, so the digit-sum per place = 3·(1 + 4 + 2 + 1) = 24. Total of all numbers = 24·1111 = 26664; mean = 26664/12 = 2222.

ModerateCAT 2022 · Slot 3TITA

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the nth day exceeds one million, then the lowest possible value of n is

Show solution

19. Since aₙ + 3 doubles each day, aₙ + 3 = 5·2ⁿ⁻¹, i.e. aₙ = 5·2ⁿ⁻¹ − 3. Requiring 5·2ⁿ⁻¹ − 3 > 1,000,000 ⇒ 2ⁿ⁻¹ > 200,000.6 ⇒ n − 1 ≥ 18 ⇒ n = 19.

ModerateCAT 2023 · Slot 1TITA

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Show solution

92. Total = 25·50 = 1250. To maximise the toppers' score, minimise the other 20: distinct integers from 30 upward, i.e. 30, 31, …, 49, summing to 790. The remaining 1250 − 790 = 460 is shared by 5 toppers ⇒ 92 each.

HardCAT 2023 · Slot 2

Let both the series a₁, a₂, a₃, … and b₁, b₂, b₃, … be in arithmetic progression such that the common differences of both the series are prime numbers. If a₅ = b₉, a₁₉ = b₁₉ and b₂ = 0, then a₁₁ equals

(1) 86
(2) 84
(3) 79
(4) 83

Show solution

(3) 79. Let the common differences be primes p (for aₙ) and q (for bₙ). a₁₉ = b₁₉ and a₅ = b₉ give relations that, with b₂ = 0, force p = 7 and q = 5. Then a₁₁ works out to 79.

ModerateCAT 2023 · Slot 2TITA

Let aₙ and bₙ be two sequences such that aₙ = 13 + 6(n−1) and bₙ = 15 + 7(n−1) for all natural numbers n. Then, the largest three-digit integer that is common to both these sequences, is

Show solution

967. The aₙ values are 13, 19, 25, … (step 6); the bₙ values are 15, 22, 29, … (step 7). Common terms step by LCM(6, 7) = 42 from the first common value 43; the largest three-digit such term is 967.

HardCAT 2023 · Slot 3

Let aₙ = 46 + 8n and bₙ = 98 + 4n be two sequences for natural numbers n ≤ 100. Then, the sum of all terms common to both the sequences is

(1) 14602
(2) 14798
(3) 15000
(4) 14900

Show solution

(4) 14900. aₙ runs 54, 62, …, 846 (step 8); bₙ runs 102, 106, …, 498 (step 4). Common terms form an AP with common difference LCM(8, 4) = 8 within the overlapping range; summing them gives 14900.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1

Suppose x₁, x₂, x₃, …, x₁₀₀ are in arithmetic progression such that x₅ = −4 and 2x₆ + 2x₉ = x₁₁ + x₁₃. Then, x₁₀₀ equals

(A) −194
(B) −196
(C) 204
(D) 206

Show solution

(A) −194. With xₙ = a + (n−1)d: 2x₆ + 2x₉ = x₁₁ + x₁₃ ⇒ 2(a+5d) + 2(a+8d) = (a+10d) + (a+12d) ⇒ 4a + 26d = 2a + 22d ⇒ a = −2d. Then x₅ = a + 4d = −2d + 4d = 2d = −4 ⇒ d = −2, a = 4. So x₁₀₀ = a + 99d = 4 + 99(−2) = −194.

HardCAT 2025 · Slot 2

Let aₙ be the nᵗʰ term of a decreasing infinite geometric progression. If a₁ + a₂ + a₃ = 52 and a₁a₂ + a₂a₃ + a₃a₁ = 624, then the sum of this geometric progression is

(A) 57
(B) 54
(C) 60
(D) 63

Show solution

(B) 54. With a₁ = a, ratio r: a(1 + r + r²) = 52 and a²r(1 + r + r²) = 624 ⇒ ar = 624/52 = 12 (dividing). Then a²r·(1+r+r²) = 624 with a(1+r+r²) = 52 gives ar = 12; combined with a(1+r+r²)=52 and a₂ = ar = 12 ⇒ a + 12 + 12r = 52 ⇒ a + 12r = 40, and a·r = 12 ⇒ a = 36, r = 1/3 (decreasing GP needs |r| < 1). Sum = a/(1 − r) = 36/(1 − 1/3) = 54.

Maxima-Minima · 17 CAT PYQs

Maxima-Minima

HardCAT 2001

Let x, y be two positive numbers such that x + y = 1. Then, the minimum value of (x + 1/x)² + (y + 1/y)² is

(1) 12
(2) 20
(3) 12.5
(4) 13.3

Show solution

(3) 12.5. By symmetry the minimum is at x = y = 1/2. Each term = (1/2 + 2)² = (5/2)² = 25/4. Sum = 2·25/4 = 25/2 = 12.5.

HardCAT 2002

For three integers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?

(1) 3√13
(2) √19
(3) 13/3
(4) √15

Show solution

(3) 13/3. Substitute z = 5 − x − y into xy + z(x+y) = 3 to get y² + (x−5)y + (x²−5x+3) = 0. For real y, discriminant ≥ 0: (x−5)² − 4(x²−5x+3) ≥ 0 ⇒ 3x² − 10x − 13 ≤ 0 ⇒ (3x−13)(x+1) ≤ 0 ⇒ −1 ≤ x ≤ 13/3. So x_max = 13/3.

HardCAT 2003

If three positive real numbers x, y, z satisfy y − x = z − y and xyz = 4, then what is the minimum possible value of y?

(1) 2^(1/3)
(2) 2^(2/3)
(3) 2^(1/4)
(4) 2^(3/4)

Show solution

(2) 2^(2/3). x, y, z are in AP so y is their AM. By AM ≥ GM, y ≥ (xyz)^(1/3) = 4^(1/3) = 2^(2/3). Minimum y = 2^(2/3).

HardCAT 2003

Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Which one of the following is necessarily true?

(1) The minimum possible value of a² + b² + c² + d² is 4m² − 2m + 1
(2) The minimum possible value of a² + b² + c² + d² is 4m² + 2m + 1
(3) The maximum possible value of a² + b² + c² + d² is 4m² − 2m + 1
(4) The maximum possible value of a² + b² + c² + d² is 4m² + 2m + 1

Show solution

(2) The minimum possible value of a² + b² + c² + d² is 4m² + 2m + 1. Sum of squares is minimised when the values are as equal as possible: three equal to m and one equal to m + 1. Then 3·m² + (m + 1)² = 4m² + 2m + 1.

HardCAT 2004

Let f(x) = ax² − b |x|, where a and b are constants. Then at x = 0, f(x) is

(1) maximized whenever a > 0, b > 0
(2) maximized whenever a > 0, b < 0
(3) minimized whenever a > 0, b > 0
(4) minimized whenever a > 0, b < 0

Show solution

(4) minimized whenever a > 0, b < 0. For a > 0, b < 0, both branches open upward with no negative dip at 0, so x = 0 is a minimum.

ModerateCAT 2006

Let f(x) = max (2x + 1, 3 − 4x), where x is any real number. Then the minimum possible value of f(x) is:

(1) 1/3
(2) 1/2
(3) 2/3
(4) 5/3

Show solution

(4) 5/3. The max of two lines is least where they cross: 2x+1 = 3−4x ⇒ x = 1/3, value = 2(1/3)+1 = 5/3.

ModerateCAT 2007

A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?

(1) −119
(2) −159
(3) −110
(4) −180

Show solution

(2) −159. Vertex form f(x) = a(x−1)² + 3 with maximum ⇒ a < 0. f(0) = a + 3 = 1 ⇒ a = −2. So f(x) = −2(x−1)² + 3 ⇒ f(10) = −2(81) + 3 = −159.

ModerateCAT 2017TITA

If a, b, c, and d are integers such that a + b + c + d = 30 then the minimum possible value of (a − b)² + (a − c)² + (a − d)² is

Show solution

2. Minimised when values are as equal as possible. 30/4 = 7.5, so take 8, 8, 7, 7. With a = 8 and b,c,d = 8,7,7: 0 + 1 + 1 = 2.

ModerateCAT 2018TITA

Let f(x) = min (2x², 52 − 5x) where x is any positive real number. Then the maximum possible value of f(x) is

Show solution

32. The min-of-two reaches its peak where they meet: 2x² = 52 − 5x ⇒ 2x²+5x−52 = 0 ⇒ x = 4. f(4) = 2·16 = 32.

ModerateCAT 2018TITA

Let f(x) = max(5x, 52 − 2x²), where x is any positive real number. Then the minimum possible value of f(x) is

Show solution

20. The max-of-two is least where they cross: 5x = 52 − 2x² ⇒ 2x²+5x−52 = 0 ⇒ x = 4. f(4) = 5·4 = 20.

HardCAT 2020TITA

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + 1/x)(1 + 1/y) is:

Show solution

2704. To minimise (1+1/x)(1+1/y) maximise x and y, so x = y = 51. Then 2601·(52/51)(52/51) = 2601·(52²/51²) = 52² = 2704.

HardCAT 2020

For real x, the maximum possible value of x/√(1 + x⁴) is:

(1) 1/√3
(2) 1/√2
(3) 1/2
(4) 1

Show solution

(2) 1/√2. For x > 0, x/√(1 + x⁴) = 1/√(1/x² + x²). By AM-GM, x² + 1/x² ≥ 2 (equality at x = 1), so the denominator is at least √2 and the expression is at most 1/√2.

ModerateCAT 2021 · Slot 3

If f(x) = x² − 7x and g(x) = x + 3, then the minimum value of f(g(x)) − 3x is

(1) − 15
(2) − 20
(3) − 16
(4) − 12

Show solution

(3) −16. f(g(x)) − 3x = (x+3)² − 7(x+3) − 3x = x² − 4x − 12. Minimum = −D/4a = −(16 + 48)/4 = −16.

ModerateCAT 2022 · Slot 1

Let 0 ≤ a ≤ x ≤ 100 and f(x) = |x − a| + |x − 100| + |x − a − 50|. Then, the maximum value of f(x) becomes 100 when a is equal to:

(1) 0
(2) 25
(3) 100
(4) 50

Show solution

(4) 50. Since a ≤ x ≤ 100, |x − a| = x − a and |x − 100| = 100 − x, so f(x) = 100 − a + |x − a − 50|. Its maximum over x ∈ [a, 100] is 100 − a + max(50, |50 − a|). This equals 100 exactly when a = 50 (giving 50 + 50 = 100).

HardCAT 2022 · Slot 3

The minimum possible value of (x² − 6x + 10)/(3 − x), for x < 3, is:

(1) −2
(2) 2
(3) 1/2
(4) −1/2

Show solution

(2) 2. Let t = 3 − x > 0. Numerator = (x−3)² + 1 = t² + 1, so expression = (t² + 1)/t = t + 1/t ≥ 2 by AM-GM (equality at t = 1, i.e. x = 2).

HardCAT 2023 · Slot 2TITA

Let k be the largest integer such that the equation (x − 1)² + 2kx + 11 = 0 has no real roots. If y is a positive real number, then the least possible value of k/(4y) + 9y is

Show solution

6. (x − 1)² + 2kx + 11 = x² + (2k − 2)x + 12. No real roots ⇒ D < 0 ⇒ (2k − 2)² − 48 < 0 ⇒ (2k − 2)² < 48, so the largest integer k = 4. Then k/(4y) + 9y = 1/y + 9y ≥ 2√(9) = 6 by AM-GM (equality at y = 1/3).

Genuinely unrecoverable / out-of-scope from this chapter's pages: a cluster of pure arithmetic word problems printed in the Algebra solved-papers but belonging to Arithmetic (cheque-transposition CAT 2000 & 2007; census Chota/Mota Hazri CAT 2001; burger-shake-fries CAT 2001; coin-change CAT 2001; Miso currency CAT 2007; rice-shop CAT 2008; bacteria-generation CAT 1998; salary X & Y CAT 2001; book page-sum CAT 2001; Fibonacci 10th term CAT 2001; telecom operators CAT 2005; cities-and-roads CAT 2000), these are covered better in the Arithmetic chapter and are not algebra-technique items. A few log/indices questions whose option text is too garbled in extraction to transcribe faithfully (CAT 1998 a#b / a∇b symbol operations Q4; CAT 1998 log M/N Q6 & Q7; CAT 1998 series log Q8; CAT 2004 log chain Q14; CAT 2005 smallest-value x=−0.5 Q12; several CAT 2017-2021 log identities printed with broken fraction glyphs) were left out rather than guessed, per the authenticity rule.

CAT 2024 & 2025, recent

ModerateCAT 2025 · Slot 2 TITA

If m and n are integers such that (m + 2n)(2m + n) = 27, then the maximum possible value of 2m − 3n is:

Show solution

17. Let m+2n and 2m+n be integer factors of 27. Their sum (m+2n)+(2m+n) = 3(m+n) must be a multiple of 3, so only the factor pairs (3, 9), (9, 3), (−3, −9), (−9, −3) qualify. Solving each gives (m, n) = (−1, 5), (5, −1), (−5, 1), (1, −5), for which 2m − 3n = −13, 13, −17, 17. The maximum is 17.